Generally an order iso-morphism or two kinds or rather two kinds of orders are order 'isomorphic' if and only if the 'order theoretic properties' of one order are precisely those of the other (trichotomy, transitivity,density) are preserved, as well as equalities, equivalence classes and top and bottom elements).

Automorphism just means "isomorphism with itself". 

So if you have an ordered structure (X,≤) an automorphism is a bijection f:X->X such that for all x,y in X, x≤y iff f(x)≤f(y).

For example, take the real numbers with the usual ordering. There are infinitely many order automorphisms of this structure - for example define f(x) = x + k, for any real number k, and you have an order automorphism; define g(x) = kx for any positive real number k, and you have an order automorphism.

And an order embedding of (X,≤) into (Y,≤) is an isomorphism of the former with a subposet (not necessarily proper) of the latter. Or, as  is said in Wiki, an isomorphism is a bijective embedding. An embedding of (X,≤) into itself could be called an endomorphism of (X,≤),

ah thanks for this; i was wondering however, in what sense must the domain and co-domain be equal when it comes to an automorphism, - does it mean one and the same elements in the domain and co-domain or that ≤ has the same meaning (outside of order theory). I presume the former. In both case the difference in meaning comes through the distinct function (ie one could have two function probabilitylike  functions that order automorphism of each other and both defined on [0,1] where x>y means greater objective probability  and F(x)>F(y) means greater subjective probability functions. Where both orders have the same order-theoretical properties (strict total bounded) linear continuum

I presume that outside of numerical domains that naturallyboth  have a strict total order (in the case of real numbers) .  whether the relation between two orders is an auto-morphism does not denote the strength of the two orders as such(neither may be total) but the fact that they have the same domain and co-domain (defined on the elements) and with the same order theoretical properties, (one is total or partial/dense /assymetric etc if the other is,) and (3)represent each other in the sense x>=y F(x)>=F(y).IN the case of numbers, A striclty monotonic increasing (but non surjective/continous) real valued function from [0,1] to [0,1] where the stronger property  x>y iff F(x)>F(y) x=y iff F(x)=F(y) where F(1)=1 F(1/2)=1/2and F(0)=0 (the same top and bottome elements and thus bounded) may not qualify as iso-morphism or a automorphism,relation between the real interval and [0,1] and codomain [0,1] or function,if not 

I presume it would be an endo-morphic order imbedding (with equal top and bottom elements) , despite [0,1] the first order being identical to itself, and the function being defined on every element of [0,1] (and both being dense strict total order with continum density that are limit  and order preserving, and  are order reflections of each other) and which strong represent each other in the sense of F(x)>F(y) iff F(x)>F(y) and in their order properties (have the same top and bottom elements). I presume that the image, may not be compact or complete  least upper bound property(and fail to be a linear continua) despite it having continuum density (perhaps it would map to all of the transcendental numbers which are uncountable in [0,1] plus zero and 1]; the least upper bound of some of these values may be rational and not in that set) Ipresume the issue is that the images are not identical , F is not invertible some elements of the co-domain do not have an element of the domain assigned to them)it fails to be an iso-morhphism because it fails to be continous or dekekent compact which could be a order theoretic property.

Would the restriction of the co-domain of F to F's image, be an  order isomorphism from [0,1] to some scattered, continnumly many yet ordered set of points  in [0,1] could be [0,1] to [0, 0.4] cup [0.6, 1] if there is a jump discontinuity(it would not be an auto-mophism as the image may be somewhat scattered, if F was non surjective/continius to begin with.

If one were to keep the original order qualitative it may qualify as an arbitary isomorphism ie in qualitive finite probability structures from A >=B iff PR(A)>=PR(B) where the first is set of events A>B>C and and is qualitative and teh second  order is an arbitary set of tthree  reals in [0,1] that sum to unity and respect PR(A)>PR(B)>PR(C) some arbitrary set of three real values that sum to one one re non-negative and respective that order,where this may not be unique yet the number of elements in the order is three in either case, (> If both order, were to remain qualitative, it may even qualify as the identity automorphism 1={A,B,C, A U B. AUC B U Cemptyset , unit < = >} to 2= {A,B,CA U B. AUC B U C empty set  unit} subject to the same qualitative axioms and both being a total (or a strict weak order with a total pre-order, with an equality relation with top and bottom elements, where A>1=B  iff A>=2 Border, (if not the identity in order terms, if both were converted into qualitative terms except a slight difference in the meaning of >=i and  >=2 (subjective probability and objective probability that is non order theoretical; does this still count as automorphism, or do these non-order theoretical terms,  mean that its a canonical isomorphism (an endomorphic isomophism where >=1 and >=2 have slightly different meaning that could not be discerned at the mathematical or order theoretical level) as qualitatively subjectively more likely and objectively more likely where the no values are assigned just the same set of events with the same order with regard to ; ie here X=Y and x>1=y iff x>2=y or is it x>=y iff x>=Fy iff F(x)>=F(y)

There could never be an order isomorphism between [0,1] and the union of [0,0.4] and [0.6,1], if both sets are ordered with the usual order as real numbers.

The reason is simple to understand. In the latter set there is a covering relation, i.e. two elements such that one immediately precedes the order - there is no element that lies strictly between 0.4 and 0.6. There is no such pair of elements in [0,1].

Think of it this way: if two structures are order-isomorphic any order property of one will also be an order property of the other. Other properties, of course, may not transfer - e.g. algebraic or topological properties might be very different, but as far as order is concerned, two order-isomorphic structures can be considered identical, safe for the actual members of each underlying set. You may substitute one for the other just by relabeling each element with its corresponding element in the isomorphism. 

For example: if (A,≤) is order-isomorphic to (B,≤') then if one has a least element so does the other, if one order is total so is the other, etc.

@ Luís Sequeira  is the requirement for an automorphism  merely that x=2 F(y)  where F is a bijection and X=Y (bijective endo-morphism order embedding) where X, >=1 and Y >=2 are one and the same ordered set (of continuumly many objects) and >=1y and >=2y may have different meaning outside of order theory (arguably it may be an identity automorphism- or this what an identity automorphism means (outside the context of number (if it makes sense to talk of cardinalities. when Both sets are finite, {A, B C} {A, B. C} any two strict total orders on that the same set that have the same relation A>B >C and A>B>C in order 2, would qualify as order automoprhism not just in the function sense as bijective automophism (as the order could be reversed in the function sense and still be a permutation or automorphism( in the order sense, if not an identity automorphism particularly  if >= does not have to means the same thing outside of order theoretical properties . I presume that the distinction between orders to not appears and between automorphisms and so forth unless one is talking about infinite structure and non strict orders) ie two total orders (that are not strict total orders ) on the same finite set, that are an order embedding may not be order automoprhic in some sense,, (depending on what = is taken to mean, in a total order,ie if one can x>1=y iff x>2=y on the same finite set, where these bijective order embeddings (both are total  order of a kind without strict total orders) on the same set of three objects one order i y>=z y>=x x>=z \neg x>=y)and the y>=z y>=x x>=z and(neg x=z (and \neg x=z (and \neg x

No, the notion of AUTOmorphism requires that X=Y and that [0,2]  (with the usual order)

  x    |---->  2x

is an order ISOmorphism between two different ordered structures 

 (Z, ≤)   ----> (Z,≤)

   x       |---->  x+1

is a bijection (between Z and Z) and also satisfies 2) as clearly x≤y iff x+1≤y+1

therefore it is an AUTOmorphism of the structure (Z,≤); obviously it is an ISOmorphism between (Z,≤) and itself.

Again, that is precisely what AUTOmorphism means: ISOmorphism with self

Ok thanks so an isomorphism requires that the relationship is (1) bijective , (2) both orders have the same rank amongst the events so mapped (though the sets may be distinct, ) which is roughly your (2) injective order embedding

and that (3) its the same order type. one is strict total , assymetric if the other is, r; however, I do not think that bijective order embedding then is suitable as requirement for a isomorphism, (3) and (1) and (2), may be satisfiable by a bijective endomorphism from   {1,2,3},>=,

plus endomorphism and the same meaning of the relation for an automorphism are met, as so is  (3) that they are the same kind of order, but they do not satisfy (2) not the same rank (not the same strict total order as the rank is distinct) so I presume it would not even be an order embedding unless restricted  much less an isomorphism or an automorphism

andnot only do they have top and bottome elements but have the same top and bottom elements, however, the rank is not the same and so (2) f(x)≤f(y)  is violated. So i presume (2) requires that not only is the same order type, but if it is,must be the same instance of that order type. if A is ranked above B then in one order then so it it in the other (which is beyond specifying the order theoretical properties but which such as totality where one order is total iff the other is, but both must be in some sense the same order amongst the events, as there are many orders that have the same order properties, on the same set, but where the order between the elements are  distinct. I presume

bijective endomorphism,

(which is distinct from (2), (3) says that one is strict total order if the other is, the first is total, assymetric if the other is, whilst (2) says the rank itself is the same (its the same strict total order, ie

even if they are the same type of order type, (total, assymetric, strict total)  , strict total order, as the rank is diametrically opposite, (strict total) yet the rank between the events in this

endomorphism is not even an embedding,  as whilst it is an endomorphism (perhaps not an order endomorphism) and whilst it is a bijectionas it has to be the same order, not merely the same order type 

the (1) bijective endomorphism and (2) order preserving and reflecting; and that the order relation means the same thing even with regard to the its non order theoretical properties (by which I mean if one relation is objective probability with the same order, and the same kind of order , if one has two identical order (same ranking )  and both strict total order on a finite, say three element set, of the same events, and both are numerical

subjective probability  in [0,1]and the other is numerical objective probability in [0,1]of three events measured by reals in [0,1](strict total)probability :

Then an automorphism would required that three reals in the domain PR, and Im are the same ie if the numbers assigned to the events are the elements (ie we identify the order elements with the numbers) and the first set is {0,1, 0,3 0, 0,6) and the other is {0.2, 0,3, 0.7}; then at best it would be an isomorphism as the two sets of numbers are distinct, disqualifying it from being an automorphism.

Does the fact that one is subjective probability and the other is objective probability ; where the elements of both order are the same X={A0.2, B0,3, C0.7, Cr,}=Y={A0.2,B0.3,C0.6, PR}

where PR(A)

@Luís Sequeira · with your second example x to x+1 if both  measured on a closed and bounded real interval [0,1] and [1,2] respectively  numbers its merely an would it be an isomorphism, both have top and bottom elements but not the same top and bottom elements as such;. however, as you have defined its an automorphism if not a stronger one (both have the not only the same order theoretical properties, not top and bottom, but they also have the same top and bottome element, trivially because of this (ie neither have one, so they trivially have the same non-existent top and bottom element). Must an endomorphic iso-morphism preserve the same order type as well as the same order, ie (must they not only have the top and bottom elements and be strict total iff the other is,

but have the same order, ie in other words not only must one be strict total if the the other is, and have top and bottom elements but the same numerically speaking top and bottom elements if endomorphic; ie 

whilst an automorphism of the unit interval might be a monotone decreasing  bijection from [0,1] to [0,1] and thus a strict monotonic decreasing ( from injection plus monotonicity)  bijection from [0,1] to [0,1] a

nd as its surjective  endomorphism as its a bijective endomorphism, endo as domain[0,1] =codomain [0,1]

and as result

domain[0,1] =codomain [0,1]=endo=image[0,1] from surjective

endo=image[0,1] its domain and image are identical and as the domain is closed and bounded, its image is likewise, and thus

F's image is an compact interval, andF  strictly monotone increasing which generally implies that  F is continuous (and over [0,1] uniformly continous with F(1)=0 and F(0)=0 from bijectivite but clearly in some sense not an order embedding (perhaps an order reversing automorphism). I presume that an order auto morphism must have the same top and bottom elements, not merely be automorphism that have the same strict order type (strictly decreasing) whilst the domain is strict increasing and so both are strict total orders arguably that are linear continua, same order type but are not the same order (they are opposite of each other) I presume the order must be the same. Whilst F(x)=x^2 is arguable an order automorphism of the unit interval as  its an automorphism and its strictly increasing,  as is the unit interval itself if were to describe the unit interval as the identity function over the unit interval the order (not order type) but the ranking itself  is the same, for instance both not only have  top and bottom elements and are both strict total order, and  bijections of the same set, but are the same order,  as in the order sense, the order is the same, and the top and bottom elements both exist, and are numerically identical 0 for F(x)=x 0 for F(x^2) and 1 for F(x) and 1 for F(x^2) as top elements

same closed bounded compact interval [0,1] so its where  and so image =[0,1]  a compact interval, istrictly decreasing (bijective)and continous function and (surjective well defined domain and codomain and

ie [0,1] to [0,1] both have top and bottom element F(x)=1-G(x) where G(x) is the identity, but these are distinct, would this event count as order embedding

Moreover, would it be an isomoprhism though if if the first were measured on the non negative integers  and the other on the positive integers, both would not be measured on the same set of have the same top and bottom elements but the order is still respected (2), is arguably bijective (1)and > means the same thing although the functions are distinct with distinct domains; both have an bottom element but not the same bottom element (0} respectively and (1) although neither have a top element . In some sense F(2x)=2F(x) is stronger perhaps a canonical isomorphism in some sense, as the bottom elements are the very same element

whilst as an the domain order 1 could be [0,1] whilst order [1,2]  they both have top and bottom elements and but not the same top and bottom elements with your second example surely it also F:G(x) to 2x where G(x) is the identity function on dom and codomain [0,1] whilst F is just F(2x) from a [0,1] to [0,2] surely it satifies (2) but neither satisfy the endomorphism requirement x\to x+1 unless measured non numerically, or or over an infinite set of categorical objects with no end point or an itial . I see you use the  entire set of integers, so neither x and x+1 have an end point or beggining pt if one point, and there is no conlict about having the same end point or intial point 

incidentally F(2x)=2F(x) satisfies this

You're making it harder than it needs to be.

Order Automorphism

is THE SAME as

Order Isomorphism when the two isomorphic structures are one and the same

is THE SAME as

an order endomorphism which is both bijective and an embedding

------ also -----

every 

I presume that any two functions that strict total orders over the very same finite set, where {A, B, C} {A, B C, >y order 1 iff x>y order 2are order preserving and relflecting, witll be identity automorphic ; presumably they could not fail to be surjective due to totality ;  and order embeddings are by definition injective ; so its an automorphism that is structure preserving with the same order type and meaning of and thus order automorphism and the same order as its an embedding; as finite it will be literally identical, and thus also the identity order automorphism.

Presumably any injective function (in the numerical sense nie single value not partial and injective, moreover injective in the function sensem) so that distinct elements of domain go to distinct elements of codomain) over a set whose domain =codomain in cardinality, will be a bijection, and if the endomorphic injective functions from finite set (say of numbers),  will be automorphic, and presumably any 2 monotone increasing function injective endo-morphism on a finite set, is the identity automorphism in both the numerical sense and order sense, where both sets  are finite, will be bijective; So I presume the literature is distinct  with regard orders, as any 2 orders which are finite endomorphisms, and are order embeddings of each other which is (monotone increasing and injective)  are not automorphic and may not be surjection, as  it appears that that the focus is on the domain, and they may be distinct order types,  and unless total, could be defined in some sense over the entire set, yet not be surjective as one order may or both orders may not be surjective ; or may not be surjections of each other ; due to a failure of totality,  would any two total orders on a finite set that

are endomorphism of each other, I presume that an one cannot say merely that the restriction of the very same total order, are endomorphic but surjective simply by sayings that A, B >)subset A, B C>  where despite the restriction , order 1, being defined on a proper subset of order 2  so that the images are distinct as IM order 1 (A, B >)s={A, B}subset {A,B.C]= Im order 2(,A B C) it arguably has the same codomain    one coulld arguably stipulate that the codomain of the restriction order 1 whose image is {A, B} is {A, B, C} as its not sur-jective which does not require that its image is {A, B C}, and despite the restriction being total, this only requires that  the restriction is total and well defined on its domain,  which is {A, B} so I presume that the distinction between domain  on the hand and image, and co-domain does not make sense when speaking about any individual order,

I

every elements of domain {A, B} not codomain or is stipulate that the codomain of the restriction  (A, B, C}    as

different image (or rather co-domain)  (a subset),, a non surjectiveand   defined on the same set, and have the same order type (both merely total orders) and theorder embeddings of each other, be the identity automorphism

be isomorphism  that is a finite endomorphism

if a an order is defined on a set and is total, it must be defined on all of its elements, (or it would not be a total ordering of that set) (order wise)

anin the sense that the set are the same,  and the meaning of the symbols are the same, and are embeddings that is over the elements on which they are defined (these orders are the very same order)  A B are the same order type (strict total) but not the same order, and thus not order embeddings, will be  the identity auto morphism

@ luis and janis must an order auto-morphism be a bijection or a bijection (and thus surjection of the order) ie consider, a non total partial strict order mapped to its numerically identical self (presumably its identical to itself and if anything is the identity auto-morphism, a function that reproduce the very same order in very aspect) is the identity, which is an automorphism and thus bijection and surjection but neither domain order, nor the function order is surjective,

I presume they need only be surj-ection of one another.  and so if a strict partial order is defined on {A, B , C, >, B & C B & C2, ie by which i mean the individual relations between the elements,are the same relation in kin not the properties of the order, ); and I presume that an isomorphism preserves not merely the the properties of least and smallest and  totality so forth AB but also if its over the same set, the very same relations if A>B, its not mapped to the inverse strict order,which has the same general order property as a strict total order with top and bottom elements, but it A itself does not stand in the same relation to B, and which elements are top and bottom elements are not preserved despite the property of having a least and greatest being preserved (ie some other element is assigned 'least and greater), i presume a many isomorphism preserve which  elements are top and bottom, and not merely (the property of having a least and greater or being a strict total order, but which relations the elements stand in, and which element is the least or greatest, is the same in some sense in both orders) not merely  the type of order, but the order relations > = < between elements are preserved, hence the name embedding or order preservation and reflection, it does not just preserve that if its a total order it is still is a total order, but the particular ordering as well?

an order embedding that it does not hold, and is distinct from injection in the sense of functions where biconditional monotonicity is roughly equivalent to strict increasing-nessand representation by a real valued ffunction with a real valued argument from [0,1] to [0,1] that is strictly increasing, generally  requires a strict total and continnum dense l order, where the domain function could be conceived as  the the identity  function,  and the strictly increasing function is a single valued real valued numerical function over that common real valued domain that is defined on every element of that domain so conceived as its order, then it will generally be considered dense and strict total if not continous in the order, despite being non-continous or non surjective (its still defined on every element of the domain, which is its order set, and in a strictly increasing and thus dense strict total order sense, particularly if F(1)=1 and F(0)=0  and even continuous in the order .

(it might only assign rationals numbers to every element of the domain [0,1] but not be a dedekind compact order) somewhat a bit like the minkowski function, buts  as its a numerical function and its order set is [0,1], as its numerical with a real valued argument and domain and injective (via strict increasing ness) its still going to a kind of strict dense total order that is continuous in the order sense (but possibly surjective or continuous not in the numerical functional analysis sense) as generally this is defined relative to the domain not the image,  particular if F(1)=1 and F(0)=0 and domain is strictly ordered by definition ; it will still have in its image, a continuum of strictly totally (and arguably densely ordered) ordered distinct points that totally ordered by strict increasing ness with top and bottom, I presume it will fail to be a a continous order complete (dedekind complete order); ie it maps everything element of the domain [0,1] to a  distinct rational numbers  in [0,1] in a strictly increasing fashion rational numbers in [0,1] like the minkowsi function.Perhaps the set of all rationals, it will be arguably separable dense a strict total order that is continous in the order  dense,separable and second countable but not dedekind complete or cmpact; there are no gaps in the order but there are holes.

As a result the function {0,1] to [0,1] is strictly increasing and the domain just is [0,1] ,or the identity function.

So if F a strictly increasing real valued function with real valued domain [0,1] and  is also conceived simultaneously   and is conceived as being an numerical order that is being strictly totally ordered by the domain=[0,1] and sdefined over the same elements as an numerical order, it is arguably a strict total order despite being possibly non surjective or possibly non continuous, its a numerical single valued function that assigns some specific number to every element in its order, the domain, in a strict order increasing and injective fashion and thus is continuous in the order, dense and strict total as well arguably  but not necessarily surjective or continous as far as numerical functions are concerned n which is numerical say the F(x)=x [0,1] to [0,1] , and is an ordeer defined as being over [0,1] not its codomain, it is a strict total order as well and defined by them, then is a strict total order  over the domain as well as well but not necessarily a surjective or continous function