If we have $H$ a normal subgroup of $G$ it is known that the conjugation action induces an action of $G$ on the cohomology $H^*(H;A)$ for any $G$-module $A$. This action is considered for the construction of the Hochshild-Serre spectral sequence. I have no problem in understanding how you can define it, but there is a basic fact that I cannot understand: that is, if $H \leq Z(G)$ then this action is trivial.
I know that the action is defined via group conjugation, so if $H$ is in the center of $G$ conjugation is trivial. But when we define the map on the cohomology groups we also have to apply some map on the coefficient module $A$.
Explicitly, if $g\in G$ then its action of $H^*(H;A)$ is defined as follows \begin{equation} H^*(H;A)\xrightarrow{res} H^*(H; \rho^*A)\xrightarrow{\mu_{g^{-1}}} H^*(H;A) \end{equation} where $\rho \colon H \rightarrow H$ is the conjugation map $x\mapsto gxg^{-1}$ and $\mu_{g^{-1}}\colon A \rightarrow A$ is just $a \mapsto g^{-1}a$ which induces a map of $H$-modules $\rho^*A\rightarrow A$. For reference I took Weibel's "An introduction to homological algebra".
As you can see we also need to operate on $A$ by multiplication by $g^{-1}$, which a priori is not trivial.
I also found as another reference this question Conjugation action group cohomology which describes the action via the cochain complex $C^*(H;A)$, but the same problem remains: using the notation of the accepted answer, the map $\phi_g$ induces the identity on the complex but we are left with $\psi_{g^{-1}}$ which is not the identity.
I would suspect that the multiplication by $g^{-1}$ on $A$ induces the identity on cohomology $H^*(H;A)$, but I could not find the proof anywhere.
This fact does not hold in general. Let $H$ be the trivial group and you can easily find a counterexample. In Brown’s book Cohomology of Groups, there is an additional condition that $A$ is a trivial $G$-module.