Recall that a permutation group $G\le\mathfrak{S}(\mathbb N)$ is oligomorphic if, for any positive integer $k$, the induced $G$-action on $k$-tuples of distinct elements of $\mathbb N$ has only finitely many orbits. Let's say that a countable group is oligomorphic if it is isomorphic (as an abstract group) to some oligomorphic permutation group.
Given a short exact sequence of (abstract) countable groups $1\to K \to G \to Q \to 1$, with $Q$ oligomorphic (or even $K$ and $Q$ oligomorphic), are there conditions on the sequence which guarantee that $G$ is oligomorphic?
EDIT: I would even be interested by the case where $K$ is finite. One motivation for me is the following. If $G$ is oligomorphic whenever there is a short exact sequence of (abstract) groups $1\to K \to G \to Q \to 1$, with $K$ finite and $Q$ countable oligomorphic, then every acylindrically hyperbolic group is oligomorphic. Indeed Hull and Osin proved in this article (ArXiv version) that every acylindrically hyperbolic group admits a highly transitive action with finite kernel.
