Let $M$ be a compact orientable manifold with finitely generated free abelian cohomology groups in even dimensions and $0$ otherwise. Conditions imposed on cohomology clearly imply that $H_i(M) \cong H^i(M)$ for any $i$ (throughout this post, (co)homology coefficients will always be integral).
Now suppose that a finite group $G$ (actually, I'm interested in the case $G = \mathbb{Z}_2$, but let's keep it a little bit more general) acts on $M$. Then both $H_i$ and $H^i$ are $G$-modules. The question is: is the action induced on cohomology necessarily the same as that induced on homology? At first I was pretty certain that this is in fact the case, because there is a $G$-isomorphism $$ H^i(M) \cong Hom(H_i(M),\mathbb{Z}) $$ coming from the universal coefficients theorem. However, even though $Hom(H_i(M),\mathbb{Z}) \cong H_i(M)$ as groups, I can't seem to prove that this also holds for $G$-modules.
Any help would be greatly appreciated.