Let $X$ be an (algebraic) K3 surface, then we have $H^{2,0}(X)=\langle \omega_X\rangle$, where $\omega_X$ is the period. Suppose $G=\langle g\rangle$ is a finite group acting on $X$ and $g$ as an automorphism of $X$ doesn't fix $\omega_X$(e.g. $g$ is purely non-symplectic), then why is $h^{2,0}(X/G)=0$?
What's in my head is $$ H^k(X,\mathbb{Q})^G\cong H^k(X/G,\mathbb{Q}). $$ by the Leray-Cartan-Serre spectral sequence. But it requires free action and it's of coefficient $\mathbb Q$. Do we have $$ H^{i,j}(X)^G\cong H^{i,j}(X/G) $$ in general?
Also, I am not sure if we always have Hodge structure on normal varieties in general. Here $X/G$ is a normal variety, I guess we can define $H^{i,j}(X/G):=H^j(X/G,\omega_{X/G}^i)$? I know the canonical divisor is well-defined for normal varieties, so for my case that $X$ is K3, $H^{2,0}(X/G)$ is well defined but is $H^j(X/G,\omega_{X/G}^i)$ well-defined in general?