**Fusion rings**

I'll more or less stick to the presentation given in this question: [1]

We define a fusion ring as follows: consider a free $\mathbb{Z}$-module $\mathbb{Z}\mathcal{B}$ with finite basis $\mathcal{B}=\{b_{1},b_{2},\cdots,b_{n}\}$. Equip this module with a binary product such that we get a $\mathbb{Z}$-algebra $\mathcal{F}=(\mathbb{Z}\mathcal{B},\cdot)$ where $b_{1}$ acts as a multiplicative identity in the ring $\mathcal{F}$ and
\begin{equation*}b_{i}\cdot b_{j}=\sum_{k}N_{ij}^{k} \ b_{k} , \quad N_{ij}^{k}\in\mathbb{N}_{0}\end{equation*} $\mathcal{F}$ is called a *fusion ring/algebra*. In particular, note that $N_{i1}^{j}=N_{1i}^{j}=\delta_{ij}$ (multiplicative identity).

We add the following bit of structure ('*invertibility*'): for every $b_{i}\in\mathcal{B}$, there exists some unique $b_{j}\in\mathcal{B}$ such that $b_{1}$ occurs in the decomposition of $b_{i}\cdot b_{j}$ and $b_{j}\cdot b_{i}$. Denote this 'inverse' by $b_{i^{*}}$. That is, \begin{equation*}\forall i \ \ \exists ! j \ : N_{ij}^{1}=N_{ji}^{1}>0\end{equation*}
where
\begin{equation*}N_{i^{*}k}^{1}=N_{ki^{*}}^{1}=\delta_{ik}\end{equation*}

**Question**

Referring back to [1], it is asserted (in the comments) that

\begin{equation*}N_{ij}^{k}=N_{j^{*}i^{*}}^{k^{*}}\end{equation*} for all $i,j,k$.

Question:For a fusion algebra with the above structure (neutrality, invertibility and associativity of product), what's the proof (or a reference for one) for this identity?

I've tried a few things but can't quite seem to get it. The result does appear to follow if we can guarantee that $*:i\mapsto i^{*}$ defines an anti-isomorphism of a fusion algebra.

**EDIT:**
After a bit of searching around, it appears that a few places seem to include "$*$" inducing an anti-isomorphism of $\mathcal{F}$ as part of the definition. I suppose that can be motivated by $b_{1}$ being guaranteed to be contained in
\begin{equation*}(b_{i}b_{j})(b_{j^{*}}b_{{i}^{*}}) \quad \text{and} \quad (b_{i}b_{j})(b_{j^{*}}b_{{i}^{*}})\end{equation*}

Still, I wonder if this is absolutely necessary...

Frobenius reciprocity: $$N_{i,j}^k = N_{i^*,k}^j = N_{k,j^*}^i.$$ In particular what you call identity II can be proved directly from the Frobenius reciprocity as follows: $$N_{i,j}^k = N_{i^*,k}^j = N_{j,k^*}^{i^*} = N_{j^*,i^*}^{k^*}.$$ So you should focus on how to prove the Frobenius reciprocity. $\endgroup$ – Sebastien Palcoux Feb 6 at 6:14