Let $G$ be a finite group and $M,N$ be two $G$ modules then $\rm{Hom}_{\mathbb{Z}}(M,N)$ is a $G$ module with action $$G\times \rm{Hom}_{\mathbb{Z}}(M,N)\rightarrow \rm{Hom}_{\mathbb{Z}}(M,N)$$ mapping $(\sigma,\eta)\mapsto \eta^{\sigma}$ where $\eta^{\sigma}(m)=\sigma\eta(\sigma^{-1}m)$.
Define Trace map to be the map $T:\rm{Hom}_{\mathbb{Z}}(M,N)\rightarrow \rm{Hom}_{\mathbb{Z}}(M,N)$ where $$\eta\mapsto\sum_{\sigma\in G }\eta^{\sigma}$$
As $\rm{Hom}_{\mathbb{Z}}(M,N)$ is a $G$ module and trace map is used left and right when defining cup products, we expect $T$ to be a $G$ module homomorphism as well. But unfortunately it turns out to be not.
For $\tau\in G$ we have $$\begin{align}T(\eta^{\tau})(m) &=\sum_{\sigma\in G}((\eta^{\tau})^{\sigma})(m) =\sum_{\sigma\in G}\sigma(\eta^{\tau})(\sigma^{-1}m) =\sum_{\sigma\in G}\sigma\tau \eta (\tau^{-1}\sigma^{-1}m)\\ \tau T(\eta)(m)&=\tau \sum_{\sigma\in G}\eta^{\sigma}(m)=\tau \sum_{\sigma\in G}\sigma\eta(\sigma^{-1}m) =\sum_{\sigma\in G}\tau\sigma \eta (\sigma^{-1}\tau^{-1}m) \end{align}$$
So, $T(\eta^{\tau})\neq \tau T(\eta)$. Is my calculation wrong or my guess is wrong ?
Both sums are the same and are equal to $$ \sum_{\theta\in G}\theta\eta(\theta^{-1}m). $$