Is it possible to find $h$ hermitian metric such that $Isom_{h}(X) \cong Aut(X)$? I suspect this is true for some class of analytic manifolds (Riemann surfaces maybe), but  my knowledge in differential geometry is very poor, so I could not conclude it. For complex manifolds, is it possible to find $h$ hermitian metric such that $Isom_{h}(X) \cong Aut(X)$ (where $Isom_{h}(X)$ means the isometries of $h$ that preserves orientation)? What are the obstructions? Any ideas on how to measure this? Is this true at least for Riemann surfaces?
Thanks in advance.
EDIT: As a friend of mine has just suggested, what about the complexification $Isom_{h}(X)^{\mathbb{C}}$? Will $Isom_{h}(X)^{\mathbb{C}} \cong Aut(X)$ for some special class of complex manifolds?
 A: The group of isometries of a compact Riemann manifold  is a  compact Lie group. Thus if  $X$ is compact and $G={\rm Aut}\;(X)$ is not compact, then  you cannot choose such a metric. On the other hand, if  $G$ is compact you can produce such a metric  by  averaging with respect to an invariant probability measure on $G$.
Edit. As Robert Bryant  indicated in his comment I have to be more careful.  If $(X,h)$ is a complex hermitian manifold, then $\DeclareMathOperator{\Isom}{Isom}$ by $\Isom_h(X)$ I understand the group of  biholomorphic maps $F: X\to X$ such that $F^*h=h$.
A: No. Check the Riemann sphere or any elliptic curve, or a punctured complex plane, or a complex plane. Otherwise yes (on Riemann surfaces), by the uniformization theorem.
A: Yes, this is so, sometimes. The prime example in the unit disc $|z|<1$ in the complex plane. This is a $1$-dimensional complex analytic manifold. Its automorphism group is
a $3$-parametric group which is isomorphic to $PSL_2(R)$.
In the unit disc there is a Riemannian metric, which is called the Poincare metric
and for which the isometries are exactly the automorphisms.
Now by the uniformization theorem, almost all Riemann surfaces have the unit disc
as the universal cover. It follows that for almost all Riemann surfaces there is such a metric, that your equality holds. It is called the hyperbolic metric. (Exceptional Riemann  surfaces are the sphere, the plane
the punctured plane and the complex $1$-parametric family of tori.)
Some of this extends to complex manifolds in higher dimension. The analog of the hyperbolic
metric is called the Kobayashi metric, and it is believed that "most" complex manifolds 
have it, though this is difficult to prove in sprecific cases. Moreover, Kobayashi metric
is frequently not a Riemannian (Hermirtian) metric but only a Finsler metric.
Much about this can be found in the books of S. Kobayashi. 
