Isometry of K3 surface.

Let $S$ be a K3 surface and $\iota$ be anti-symplectic involution of $S$. Suppose that $g$ is a Kahler-Einstein metric on $S$. My question is;

Why $\iota$ is an isometry of $S$ with respect to $g$? Is this true for any holomorphic action of $S$?

Edit $\iota$ is called anti-symplectic if it acts on $\Omega^{2,0}$ as $-id$.

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Does 'anti-symplectic involution of $S$' mean a holomorphic involution of $S$ that carries a holomorphic volume form on $S$ to its negative? –  Robert Bryant Dec 30 '12 at 15:07
I don't understand the question. Are you implicitly assuming that $g$ is invariant under $\iota$? Why would the choice of complex structure affect whether a map is an isometry? –  Johannes Nordström Dec 30 '12 at 15:12
@Robert Yes, it does. I added the definition. –  Zheng Dec 30 '12 at 15:26
@Johannes I am not assuming that $g$ is invariant under $\iota$. As to the second question, you are right. $g$ is Kahler-Einstein for any complex structure obtained by hyperKahler rotation but $\iota$ is not necessarily holomorphic in other complex structure. –  Zheng Dec 30 '12 at 15:31
@Johannes So the question should be "Assume $\iota$ is anti-symplectic for some complex structure, then is it isometry?". But I simplified my question above. Thank you for pointing out this. –  Zheng Dec 30 '12 at 15:33

There is a unique Ricci-flat Kähler metric in each Kähler class of $S$. Thus, for any holomorphic automorphism $\iota$ of $S$, a Ricci-flat Kähler metric $g$ is invariant under $\iota$ if and only if its Kähler class $[\omega_g] \in H^{1,1}(S)$ is.
One can average the Kahler form and get invariant Kahler class. For such a Kahler metric, $g$ is invariant and thus $\iota$ is an isometry. This works for any holomorphic $G$-action. –  Zheng Dec 30 '12 at 23:04
Sure, that tells you that for any automorphism of $S$ (of finite order at least) there exists some invariant Kähler class, and hence an invariant Ricci-flat Kähler metric. But it does not mean that a given Ricci-flat Kähler metric is invariant, which is what your question seems to ask. –  Johannes Nordström Dec 30 '12 at 23:36