Byott's insolvability conjecture for realizable group pairs

Let $G$ and $N$ be finite groups of the same order. The pair $(G,N)$ is realizable if there exists a regular subgroup of $\operatorname{Hol}(N)$ isomorphic to $G$; equivalently, there is a skew brace with additive group isomorphic to $N$ and multiplicative group isomorphic to $G$.

Byott's conjecture. If $(G,N)$ is realizable and $G$ is insolvable, then $N$ is also insolvable.

This conjecture concerns restrictions on the group types of Hopf–Galois structures and skew braces. The converse is false, so realizability does not imply that $G$ is insolvable whenever $N$ is.