Let $(N \subset M)$ be a finite index inclusion of ${\rm II}_1$ factors, and $N \subset M \subset M_1$ the basic construction.
The subfactor $(N \subset M)$ is called self-dual if it is isomorphic to its dual $(M \subset M_1)$.

Let $R$ be the hyperfinite ${\rm II}_1$ factor. We will use the outer action of any finite group $G$ on $R$, and the fixed point subfactor $R^G \subset R$.

Remark: Let $G$ be a finite group. Then, $(R ^ {G} \subset R)$ is self-dual iff $G$ is self-dual, iff $G$ is abelian.

Consider the group-subgroup subfactors of the form $(R ^ {G \rtimes H} \subset R^H)$, with $G$ a finite abelian group and $H$ a subgroup of $Aut(G)$.

Question: Is a self-dual group-subgroup subfactor necessarily of this form? Is it also sufficient?

Remark: It is true at index $3$ and $5$ (see here p37). I've to check index $4$.