Let $G$ be a group and $V$ be a $G$-module. Suppose that there exists a finite group $\Gamma$ acting on both $G$ and $V$ in a compatible way, in the sense that \begin{equation} \gamma(g.v) = \gamma(g).\gamma(v) \qquad \forall g \in G, v \in V, \gamma \in \Gamma. \end{equation} If you prefer, this corresponds to $V$ being a module for the skew group ring of $\Gamma$ acting on $G$.

**Is there any relation between the group cohomology $H^*(G,V)$ and the cohomology $H^* \left( G^{\Gamma}, V^{\Gamma} \right)$ of the fixed subgroup $G^{\Gamma}$ with coefficients in the fixed submodule $V^{\Gamma}$?**

I have been purposefully vague on both the coefficient ring over which this happen (i.e. if $V$ is a complex vector space, or just an abelian group...), and the nature of the group $G$ (finite group, profinite group, topological group... - in the last two cases I mean continuous cohomology for the continuous module $V$) because I am happy with answers in any of these cases.

Some minor thoughts: obviously $\Gamma$ acts on $H^*(G, V)$, but I think it's too much to hope that $H^*(G,V)^{\Gamma}$ is isomorphic to $H^* \left( G^{\Gamma}, V^{\Gamma} \right)$. On the other hand the universal coefficient theorem says that (under some assumptions on, say, $H^*(G,V)$) taking $\Gamma$-invariants `commutes' with taking the homology of the complex computing $H^*(G,V)$ - let me denote this complex by $C^{\bullet}(G,V)$.

Suppose now that $C^{\bullet} (G,V)$ comes from the bar resolution. Then the map `restrict to $G^{\Gamma}$' from $C^{\bullet}(G,V)^{\Gamma}$ to $C^{\bullet} \left( G^{\Gamma}, V^{\Gamma} \right)$ makes sense. This map respects the differentials, so taking homology the two complexes, gives us $H^*(G,V)^{\Gamma} \rightarrow H^* \left( G^{\Gamma}, V^{\Gamma} \right)$. Can we say anything about this map?