Suppose I have a group extension $1 \rightarrow N \rightarrow H\rightarrow K\rightarrow 1$, and we have a group $G$ which acts on $H$, and $K$ by automorphisms and it does not have action on $N$. Assume that the group action is compatible with the short exact sequence. That means if $f$ is the surjective map between $H$ and $K$, we require the group action to be equivariant: $$g \,\triangleright_K\,f(h)=f(g\,\triangleright_H h),$$ where $g\in G$, $h\in H$, and the symbol $\triangleright$ represents the group action. If I consider the group cohomology with coefficient in $\mathbb{Z}$ with trivial group action, the group cohomology of $H$ will be given by those of $N$ and $K$ from spectral sequences up to an extension problem. The group action of $G$ on $N$, $H$, and $K$ will induce actions of the group $G$ on the respective group cohomologies. Assume that the spectral sequence stabilizes on the second page and there is no extension problem. My question is that: is it true that the group action of $G$ on the group cohomology of $H$ can be given by the group action of $G$ on $E^{p,q}_2=H^p(K,H^q(N,\mathbb{Z})$. That is to say: does the group $G$ act "equivariantly" with respect to the "inverse of the filtration"?
My question is related to a previous question (Spectral sequence for equivariant group extension). However, I do not understand the explanation in the comment there.