If $f:X\to Y$ is a finite morphism, and assuming that both $X$ and $Y$ admit dualizing sheaves, then by duality (see [Hartshorne, Ex.III.6.10]) you have the first map of the following:
$$
\eta: f_*\omega_X\to \mathscr Hom _Y(f_*\mathscr O_X,\omega_Y)\to \omega_Y,
$$
where the second map is induced by the natural map
$$
\varrho: \mathscr O_Y \to f_*\mathscr O_X.
$$
Now, if $f$ is the quotient map by a finite group $G$, then $\varrho$ is just the embedding of the set of $G$ invariant sections of $f_*\mathscr O_X$.
Similarly, in that case $\eta$ is $G$-invariant and I am guessing that it is the largest $G$-equivariant quotient of $f_*\omega_X$ on which the induced $G$ action is trivial. I suppose this should tell you what the quotient is, but the problem is that this map is not necessarily surjective, the cokernel is contained in
$$
\mathscr Ext^1_Y\left( f_*\mathscr O_X\big/\mathscr O_Y, \omega_Y\right).
$$
Of course if you can prove that this is trivial, then you're in business. It does happen frequently that $f_*\mathscr O_X$ is free and $\mathscr O_Y$ is a direct summand in which case the above $\mathscr Ext^1$ is zero. But in general this will depend on $G$, the singularities of $X$ and the action.
