Is there a noncommutative simple fusion ring?

A fusion ring $$\mathcal{F}$$ is given by a finite set $$B = \{b_1,b_2, \dots, b_r \}$$ such that $$b_i b_j = \sum_k n_{i,j}^k b_k$$ with $$n_{i,j}^k \in \mathbb{Z}_{\ge 0}$$, satisfying axioms slightly augmenting the group axioms (see the details here).

The fusion ring $$\mathcal{F}$$ is called noncommutative if $$\exists i,j$$ with $$b_ib_j\neq b_jb_i$$. A fusion subring is given by a subset $$B_S=\{b_s \ | \ s \in S\} \subseteq B$$ with $$1 \in S = S^*$$ and $$\forall i,j \in S$$ then $$n_{i,j}^k \neq 0$$ only if $$k \in S$$. It is called simple when every fusion subring is given by the subset $$\{b_1\}$$ or $$B$$. Let $$G$$ be a finite group, then the Grothendieck ring of $$Rep(G)$$ is simple (as fusion ring) if and only if $$G$$ is simple.

Question: Is there a noncommutative simple fusion ring?

Investigation

According to this post, a noncommutative fusion ring must be of rank at least $$5$$, and at rank exactly $$5$$, we can assume: $$2^*=3$$, $$4^*=4$$ and $$5^*=5$$. A brute-force computation shows that at multiplicity at most three, there is a unique simple example (upto equiv.):
$$\begin{smallmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\end{smallmatrix} , \ \begin{smallmatrix}0&1&0&0&0\\0&0&1&1&0\\1&0&0&0&1\\0&0&1&0&1\\0&1&0&1&1\end{smallmatrix} , \ \begin{smallmatrix}0&0&1&0&0\\1&0&0&0&1\\0&1&0&1&0\\0&1&0&0&1\\0&0&1&1&1\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&1&0\\0&0&1&0&1\\0&1&0&0&1\\1&0&0&1&1\\0&1&1&1&1\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&1\\0&1&0&1&1\\0&0&1&1&1\\0&1&1&1&1\\1&1&1&1&2\end{smallmatrix}$$ but it is commutative. Note that $$[d(b_1),d(b_2),d(b_3),d(b_4),d(b_5)] =[1,\alpha,\alpha,\beta,\gamma]$$ with:

• $$\alpha =1+2cos(2\pi/7) \simeq 2.246979603\dots$$,
• $$\beta =1-2cos(6\pi/7) \simeq 2.801937735\dots$$,
• $$\gamma =\alpha+\beta-1 \simeq 4.048917339\dots$$,
• so that: $$gdim:=\sum_i d(b_i)^2 \simeq 36.65039990\dots$$,

moreover, it is of Frobenius type and satisfies Schur Product Property.

The fusion ring of one of the even parts of Extended Haagerup will work (see Appendix A of our paper). It has rank 8, and you can see that $$AB \neq BA$$. Surely there's less complicated examples though.
$$\begin{smallmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{smallmatrix} , \ \begin{smallmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \end{smallmatrix} , \ \begin{smallmatrix} 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \end{smallmatrix} , \ \begin{smallmatrix} 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 & 1 & 1 \end{smallmatrix},$$ $$\begin{smallmatrix} 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 2 & 2 & 2 \\ 0 & 0 & 0 & 1 & 1 & 2 & 2 & 2 \\ 0 & 1 & 1 & 1 & 1 & 2 & 2 & 2 \end{smallmatrix} , \ \begin{smallmatrix} 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 2 & 2 & 2 \\ 1 & 1 & 1 & 1 & 2 & 4 & 3 & 3 \\ 0 & 1 & 1 & 1 & 2 & 3 & 3 & 4 \\ 0 & 1 & 1 & 1 & 2 & 3 & 4 & 4 \end{smallmatrix} , \ \begin{smallmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 2 & 2 & 2 \\ 0 & 1 & 1 & 1 & 2 & 3 & 3 & 4 \\ 1 & 1 & 1 & 1 & 2 & 3 & 4 & 3 \\ 0 & 1 & 1 & 1 & 2 & 4 & 3 & 4 \end{smallmatrix} , \ \begin{smallmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 & 2 & 2 & 2 \\ 0 & 1 & 1 & 1 & 2 & 3 & 4 & 4 \\ 0 & 1 & 1 & 1 & 2 & 4 & 3 & 4 \\ 1 & 1 & 1 & 1 & 2 & 4 & 4 & 4 \end{smallmatrix}.$$
It is a rank $$8$$ noncommutative simple fusion ring. If $$\mathcal{B}$$ is its basis then $$\mathbb{C}\mathcal{B}=\mathbb{C}^4 \oplus M_2(\mathbb{C})$$.