I have found that this function must satisfy another known functional equation : f(x+y)=f(x)f(y)/f(x)+f(y) which is itself reduced to the well known Cauch functional equation g(x+y)=g(x)+g(y) where g(x) = 1/f(x)
If you do not require f to be continuous, then you can find many functions satisfying f(f(x)) = x and f(1) = c. Disregarding continuity, how can you adjust your example to define a new function with the desired properties?
Suppose f must be continuous. If c = 1 then defining f(x) = x will work. If c = -1 then defining f(x) = -x will work. Similar examples also exist for an arbitrary c. Even when supposing continuity, the function f need not be unique.
Keyword: involution
Take any decreasing bijective function g:]0,1] \to [1, \infty[, for example 1/x^\alpha, and define f(x)=g(x) if 0
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