If a group G preserves the Riemannian metric, it a fortiori, preserves the conformal class of the metric and so it acts holomorphically or antiholomorphically on the Riemann surface C with complex structure defined by the conformal class of the metric, depending on whether it acts orientation preserving or orientation reversing. Let G_0 be the subgroup of orientation preserving automorphisms. Then fixpoints of G_0 are in particular ramification points  of the map C \to D = C/G_0.  

Thus, all we have to do is to take a group of orientation preserving automorphisms and avoid having ramification. But that is easy: just start with a curve D of genus g(D) > 2 and take an unramified covering of some degree d > 0. There are plenty unramified coverings because they are classified by the fundamental group which is well known to be generated by generators \alpha_i, \beta_i, i = 1,...,g with one relation  \prod_i [\alpha_i, \beta_i] = e.  Moreover the covering transformations of C \to D, being holomorphic transformation lift to the tangent bundle TC, and if we just pull back a metric from D, they automatically preserve the inner product.

Dear Waldemar Thank you very much for  your encouragement.

Dear Rogier Thank you very  much for your  answer. Can I ask you elaborate your answer. 

Some  questions  on the first part of  your answer. 

For the moment we replace the  The  Riemann surface  M_g by  \mathbb{C}

A  group  G can act on \mathbb{C}  in a  non holomorphic  or anti holomorphicn   manner but can act trivialy  on the trivial  tangent space of  \mathbb{C}, so the action preserves the inner product. Then the equivariant structure preserves the inner product of the fibers but its  action on the basew space is not holomorphic or  anti holomorphic. Am I missing some thing?

I would  appreciate you expand your answer  assuming i have  no  back ground in algebraic geometry.

Best regards

Ali

@Ali Taghavi. Mmm you have a point there. A trivial group action is a silly counterexample but a counterexample nonetheless.  

Come to think of it, all I really  do is point out that Riemann surfaces exist admitting a group of differentiable, transformations, that act fixpoint free,  that with the standard action on the tangent bundle they act in a metric preserving way if you pull up your metric from below, and if you assume they are orientation preserving, they are in fact holomorphic for the complex structure induced by the metric (hence the group acting on the Riemann Surface is finite in fact bounded by 42*(2g - 2), by Hurwitz theorem but that does not help you).

MMm let me think about this some more before I say something stupid.