Let $X$ be a finite set, $G$ a finite group and $M$ another Abelian (multiplicative) group. Let us have a transitive (left) action $G \times X \to X$ and an action $G \times M \to M$ by automorphisms. Then we have the right action of $G$ on the set of maps ${\rm Map}(X,M)$: for $\sigma: X \to M$, $g \in G$, $x \in X$ define $\sigma^g (x) = g^{-1}\sigma(gx)$. If $\sigma^g = \sigma$ then $\sigma$ is called a $G$-equivariant map.

Furthermore, let $G$ be a semidirect product of a normal subgroup $N$ and a subgroup $K$, and $\lambda: X \to M$ is a $K$-equivariant map. It is easy to see that in this case the map $$ \sigma(x)=\prod_{a \in N} \lambda^a (x) $$ is $G$-equivariant.

The converse is true under some conditions. E. g.:

Let $G=N \leftthreetimes K$ and for some $x \in X$ its stabilizer is contained in $K$. Then for every $G$-equivariant map $\sigma$ there is such a $K$-equivariant map $\lambda: X \to M$ that $\sigma$ is a product as above.

QUESTION: Does whoever know some weaker (or similar) conditions under which $G$-equivariant $\sigma$ factors by a $K$-equivariant $\lambda$?

This question arises in studying partial actions of groups.

Thank you in advance.