Given a group extension:
$$ 0 \rightarrow N \rightarrow G \rightarrow \frac{G}{N} \rightarrow 0 $$
I need to show that the induced action of $G$ by conjugation is trivial on the Hochschild-Serre spectral sequence.
$\textbf{Attempt 1}$: So far from brown I know that given a $G$-module M, we can have $G$ acting by conjugation, for a $g \in G$ we get: $$ (\phi_g, \alpha_g) : (G,M) \rightarrow (G,M) $$ acting with $\phi_g(x) := gxg^{-1}$ and $\alpha_g(m) := gm$. We know this induces the trivial map on $H_n(G, M)$ but I need to show it induces the trivial map on $$ E_{i,j}^2 = H_i(G/N, H_j(N, M))$$ which I've been really struggling to do. I know that to induce a map on this homology I need a pair of maps $$ (\psi_g, \beta_g) : (G/N, H_j(N, M)) \rightarrow (G/N, H_j(N,M))$$ and I guess that $\psi_g$ will be conjugation as above. But I dont know how to interpret the map $\beta_g$. I know it also needs the property that $\beta_g(ab) = \psi_g(a)\beta_g(b)$ for $a \in G/N$ and $b \in H_j(N,M)$. I also feel like using the usual bar resolution should make this easier to calculate by hand at some later stage but I just cant make progress at the moment.
$\textbf{Attempt 2}$: An alternate route I thought was that $g \in G$ gives an automorphism of the above group extension, which then of course extends to the long exact sequence on homology and then from there this is what we can use to get the exact couple that hochschild-serre comes from but this is only the 1'st page and it seems incredibly messy to track it then through to the derived couple.
Anyone have any ideas how to show this induced action is trivial?
We know for any group $G$ and $M$ a $G$-module then $G$ acts by conjugation on iteself and $g\cdot m$ on $M$ induces the trivial action on $H_n(G, M)$. [see Browns cohomology of groups book]
In particular since we can make $H_j(N,M)$ into a $G/N$-module by this same action when $N\trianglelefteq G$ we get by the first point that $G/N$ acts trivially on $H_i(G/N, H_j(N, M))$ by the action described above (conjugation and left multiplication).
Hence for a $g \in G$ we can map $G \rightarrow G/N$ and then act on $H_i(G/N, H_j(N,M))$ in order to get an action of $G$ on $H_i(G/N, H_j(N,M))$ which is trivial since the $G/N$ action was trivial.