This is inspired by https://math.stackexchange.com/a/94557/214353 where a (presumably standard and classical) pattern is described. Second cohomology group $H^2(G;M)$ of a group $G$ with coefficients in a $G$-module $M$ can be used to classify group extensions $p:X\twoheadrightarrow G$ with a choice of an isomorphism between $\ker(p)$ and $M$ that matches the original $G$-action on $M$ with the action arising from $p$. The class of $p$ is trivial iff $p$ admits a splitting, i. e. a homomorphism $s:G\to X$ with $ps=\mathrm{id}_G$, and in this case the set of all such sections modulo the action of $M$ by conjugation in $X$ can be identified with $H^1(G;M)$, using the fact that such an $s$ defines a crossed homomorphism $G\to M$ and conjugations by $M$ correspond to adding crossed homomorphisms of the form $g\mapsto gm-m$ for some fixed $m\in M$.
My question is about the next level. It is by now common knowledge that $H^3(G;M)$ can be used to classify monoidal functors $P:(\mathscr X,\otimes,I)\twoheadrightarrow\operatorname{disc}(G)$ surjective on objects, where $(\mathscr X,\otimes,I)$ is a monoidal category with every object $\otimes$-invertible up to isomorphism, together with a choice of isomorphism between $\operatorname{Aut}_{\mathscr X}(I)$ and $M$ that matches the original $G$-action on $M$ with the action arising from $P$. The class of $P$ is trivial iff $P$ admits a splitting, i. e. a monoidal functor $S:\operatorname{disc}(G)\to\mathscr X$ together with a choice of an isomorphism between $PS$ and the identity functor of $\operatorname{disc}(G)$.
And now I want to continue and say that the collection of all such $S$ modulo something can be identified with the set of extensions of $G$ by $M$ as above modulo extension equivalence. How does it go?
