Apologies for what may very well be a trivial question from a non AG person. Suppose I have a morphism of varieties $f: X\rightarrow Y$, with $Y$ affine, which is equivariant with respect to the action of a compact Lie group G. Suppose I know that $\pi_{*}\mathcal{O}_X \simeq\mathcal{O}_Y$ (in my case Y is a rational singularity), and \begin{equation} \Gamma(X, \mathcal{O}_X) = H^0(X, \mathcal{O}_X) = H^0(Y, \pi_*\mathcal{O}_X) \simeq \Gamma(Y, \mathcal{O}_Y) \end{equation}
Is it true then that $\Gamma(X, \mathcal{O}_X)$ and $\Gamma(Y, \mathcal{O}_Y)$ are isomorphic as $G$ modules?
My issue is I don't know what 'form' the isomorphism $\pi_{*}\mathcal{O}_X \simeq\mathcal{O}_Y$ takes, and so don't know if I can use the equivariance of the map $\pi$. Given the above isomorphism what is the way to obtain sections of $\mathcal{O}_Y$ from those of $\mathcal{O}_X$?
I'd also like to know if the same is true for the canonical sheaf, i.e. if the statement is also true replacing $\mathcal{O}_X$ with $\omega_X$ in all of the above.
On $U \subset Y$ open, we have $\mathcal \pi_*\mathcal O_X(U) = \mathcal O_X(\pi^{-1}(U))$. So when one has a function $f : U \to \Bbb A^1$ we can take its pullback $f \circ \pi \in \pi_* \mathcal O_X(U)$. Pullback commutes with restriction and hence define a map of sheaves $ \mathcal O_Y \to \pi_* \mathcal O_X$.
So unless stated otherwise, $\mathcal O_Y \cong\pi_* \mathcal O_X$ means specifically that the previous map is an isomorphism. In your case, it is a $G$-equivariant map so indeed the global sections are isomorphic as $G$-modules.
However in the setting of algebraic geometry it is more commun to have a reductive group rather than a compact Lie group (for example, the complexification of your compact Lie group if you look over $k = \Bbb C$).
The answer is the same with $\mathcal O_X$ replaced by $\omega_X$.