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.   

Let $G$ be a finite abelian group and $H$ a subgroup of $Aut(G)$.    

*Question*: Is every self-dual group-subgroup subfactor of the form  $(R ^ {G \rtimes H} \subset R^H)$?    

*Remark*: It is true at index $3$ and $5$ (see [here][1] p37). I've to check index $4$.     

*Remark (after an exchange with Marcel Bischoff)*: The question above asks if this form is necessary. We know that this form is not sufficient, because $S_4 = A_4 \rtimes \langle (1,2) \rangle$, and the double coset algebra for the inclusion $(\langle (1,2) \rangle \subset S_4)$ is not commutative (see [here][2] p45). Now the subfactor planar algebra $P=P(N \subset M)$ is selfdual only if $N' \cap M_1 = P _{2,+} \simeq P_{2,-} = M' \cap M_2$, but for this example, one is commutative and the other is not, so this $(N \subset M)$ is not self-dual.


  [1]: http://arxiv.org/pdf/1304.6141v2.pdf
  [2]: http://arxiv.org/pdf/1505.06649v9.pdf