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.

@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.

@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.

@Robert Yes, it does. I added the definition.