This is Example $6.7.10$ in Weibel's "An introduction to homological algebra", after Example $6.7.7$ in which Weibel explained how does $G$ act by "conjugation" on (co)homologies of its normal subgroup with coefficients in a $G$-module $A$.
Example $6.7.10$ (Dihedral groups) The cyclic group $C_m$ is a normal subgroup of the dihedral group $D_m$, and $D_m/C_m\cong C_2$. To determine the action of $C_2$ on the homology of $C_m$, note that there is an element $g$ of $D_m$ such that $g\sigma g^{-1} = \sigma^{-1}$. Let $\rho: C_m \to C_m$ be conjugation by $g$. If $P$ denotes the $(\sigma-1,N)$ complex of $6.2.1$, consider the following map from $P$ to $\rho^\#P$:
An easy calculation shows that the map induced from conjugation by $g$ is multiplication by $(-1)^i$ on $H_{2i-1}(C_m;\mathbb{Z})$ and $H^{2i}(C_m;\mathbb{Z})$.
I am sure that in the diagram, $G$ equals $C_m$ (because if $G=D_m$, then the horizontal sequence is not exact at $\mathbb{Z} G$). Weibel seems to suggest that after tensoring the horizontal sequences with $\mathbb{Z}$ over $\mathbb{Z}[C_m]$, the vertical arrows induce maps between homologies.
Question. The vertical maps, defined as multiplication by $(-\sigma)^i$ for appropriate $i$, are not $C_m$-module homomorphisms (because $\sigma$ acts by $g\sigma g^{-1}=\sigma^{-1}$ on the bottom copy of $\mathbb{Z}G$), but we can tensor it with the trivial $\mathbb{Z}$ to get a well-defined map $(-\sigma)^i\otimes 1: \mathbb{Z}[G]\otimes_{\mathbb{Z}[G]} \mathbb{Z} \to \mathbb{Z}[G]\otimes_{\mathbb{Z}[G]} \mathbb{Z}$. Could anybody explain why Weibel considers this diagram (in particular, this non-$C_m$-equivariant chain map) and how this helps to determine the action of $C_2$ on the homologies? Thanks in advance.
The bottom sequence is a translation of the $(g\sigma g^{-1} = \sigma^{-1},N)$ complex for the group $\langle \sigma^{-1} \rangle \cong C_m$ into the language of modules over $\langle \sigma\rangle\cong C_m$ along the homomorphism $\rho$. So, the vertical arrows are $\langle \sigma\rangle$-linear.
To verify that tensoring/hom-ing the diagram with $\mathbb{Z}$ gives the conjugation action, it suffices to understand that they somehow give rise to the unique natural transformations (between the $\delta$-functors) -- namely, coresctriction and restriction. Note that since $\rho: \langle \sigma \rangle \to \langle \sigma^{-1} \rangle, \sigma\mapsto \sigma^{-1}$ is a group isomorphism, it induces an isomorphism of rings $\mathbb{Z}\langle \sigma \rangle \to \mathbb{Z}\langle \sigma^{-1} \rangle$, still denoted by $\rho$.
For any $\langle \sigma^{-1} \rangle$-module $A$, we have the natural isomorphism of abelian groups $$ \mathbb{Z}\langle \sigma \rangle \otimes_{\mathbb{Z}\langle \sigma \rangle} \rho^\# A \xrightarrow{\rho \otimes \mathbb{1}_A} \mathbb{Z}\langle \sigma^{-1} \rangle \otimes_{\mathbb{Z}\langle \sigma^{-1} \rangle} A, $$ where the right-hand side computes $H_\bullet(\langle \sigma^{-1} \rangle;A)$.
So we have an explicit description of corestriction $$ \mathrm{cor}_{\langle \sigma \rangle}^{\langle \sigma^{-1} \rangle}: H_\bullet(\langle \sigma \rangle; \rho^\# A) \longrightarrow H_\bullet(\langle \sigma^{-1} \rangle; A). $$
For the restriction map between cohomologies, we consider the natural isomorphism of abelian groups $$ \hom_{\langle \sigma^{-1} \rangle}(\mathbb{Z}\langle \sigma^{-1} \rangle, A) \xrightarrow{\rho^* \circ (\mathbb{1}_A)_{*}} \hom_{\langle \sigma \rangle} (\mathbb{Z}\langle \sigma \rangle, \rho^\# A), $$ where the right-hand side computes $H^\bullet (\langle \sigma \rangle; \rho^\# A)$.
So we have an explicit description of restriction $$ \mathrm{res}_{\langle \sigma \rangle}^{\langle \sigma^{-1} \rangle}: H^\bullet(\langle \sigma^{-1} \rangle; A) \longrightarrow H^\bullet(\langle \sigma \rangle; \rho^\# A). $$