Consider the usual functor that associates to each topological space its singular chains \begin{align} \textbf{Top} &\to \text{Ch}(\textbf{Ab}) \\ X &\mapsto C_{\bullet}(X). \end{align}
If we consider topological spaces with an action of a (discrete) group $G$, then this functoriality above allows us to have a functor \begin{align} G\textbf{-Top} &\to \text{Ch}(G\textbf{-mod}) \\ X &\mapsto C_{\bullet}(X), \end{align} and the homology of this chain complex gives the usual homology of the space $X$, together with the group action that it inherits from the one on $X$. My question is about the proper way to do this in cohomology instead, because there are two non-equivalent ways of dualizing:
On the one hand we could dualize the abelian groups, and get a $G$-action by premultiplication $$ C^{\bullet}(X) \stackrel{\text{def}}{=} \text{Hom}_{\textbf{Ab}}(C_{\bullet}(X), \mathbb{Z}), \qquad (g \cdot \varphi)(\sigma) \stackrel{\text{def}}{=} \varphi(g \cdot \sigma). $$
On the other hand we could directly consider $C_{\bullet}(X)$ as a $\mathbb{Z}G$-module and dualize there $$ C^{\bullet}(X) \stackrel{\text{def}}{=} \text{Hom}_{\mathbb{Z}G\textbf{-mod}}(C_{\bullet}(X), \mathbb{Z}G). $$
The first definition seems to be the correct one, if we wish that the cohomology of the cochain complex $C^{\bullet}(X)$ gives the usual cohomology $H^\bullet(X)$ with the induced $G$-action. This is because the abelian groups you consider in that case are exactly the same as if there was not a group action. Hence my question is: what would you get if you chose the other definition? Can it be expressed in terms of the usual singular cohomology?
