Let $G$ be a topological group.(You can assume it is finite.) Let $A$ be a $G-$module which is also a topological abelian group equipped with $G-$action. Denote $\mathcal{C}^n(G,A)=\{f:G^{n}\to A\vert f$ continuous$\}$ and $C^n(G,A)=\{f:G^{n+1}\to A\vert f$ is $G-$module morphism$\}$.
Define the map $\mathcal{C}^n(G,A)\to C^n(G,A)$ by $f(\sigma_1,\dots,\sigma_n)\to \sigma_0 f(\sigma_0^{-1}\sigma_1,\dots,\sigma_{n-1}^{-1}\sigma_n)$. I have verified the inverse map is $G-$module morphism. I have trouble to verify this map is a $G-$module morphism.
The reason that I need the $G-$module morphism here is that I want to obtain the differential map being $G-$module morphism directly by identification. Note that $\mathcal{C}^{\cdot}(G,A)$ and $C^n(G,A)$ both induce cohomology and the cohomological group can be identified. $C^n(G,A)$'s differential is naturally $G-$module morphism. If I can show above isomorphism is $G-$module map, then I have obtained natural differential as $G-$module map.
$\textbf{Q:}$ I am not sure about $G-$module structure defined $\mathcal{C}^n(G,A)$, though the book says it is left action defined by $\sigma\cdot f(x_1,\dots, x_n)=\sigma f(\sigma^{-1}x_1,\dots,\sigma^{-1}x_n)$. Denote $\psi:\mathcal{C}^n(G,A)\to C^n(G,A)$. Take $g\in G$. Then $\psi(g\cdot f)=\sigma_0gf(g^{-1}\sigma_0^{-1}\sigma_1,\dots, g^{-1}\sigma_{n-1}\sigma_n)$. I could not match it with $g\psi(f)=g\sigma_0 f(\sigma_0^{-1}\sigma_1,\dots,\sigma_{n-1}^{-1}\sigma_n)$. How to verify $\psi$ is a $G-$module morphism?
Reference: Neukirch Cohomology of Number Fields Chapter 1 Sec 2