Positive system of algebraic integers Let $\mathbb{A}$ be the ring of algebraic integers. Consider a sequence $(d_i)_{i \in I}$, with $I$ a finite set and $d_i \in \mathbb{A} \cap \mathbb{R}_{\ge 1}$, such that $$d_i d_j = \sum_{k \in I} n_{i,j,k} d_k,$$ for all $i,j \in I$, with $n_{i,j,k} \in \mathbb{Z}_{\ge 0}$.
A subset $J \subset I$ is called a subsystem if $\forall i,j \in J$ and $\forall k \not \in J$ then $n_{i,j,k} = 0$.   Let $\langle i \rangle$ be the smallest subsystem containing $i$.
Question: Let $i,j \in I$ such that $\langle i \rangle = \langle j \rangle$. Is it true that $\mathbb{Z}[d_i] = \mathbb{Z}[d_j]$?
Observation: The vector $v = (d_k)_{k \in I}$ is a common eigenvector for the matrices $M_i = (n_{i,j,k})_{j,k \in I}$, with eigenvalue $d_i$, because: $$ M_i v = (\sum_{k \in I} n_{i,j,k} d_k)_{j \in I} = (d_i d_j)_{j \in I} = d_i v.$$ By Frobenius-Perron Theorem (stated below), the eigenvalue $d_i$ of $M_i$ is its spectral radius.
Frobenius-Perron Theorem: A square matrix $M$ with nonnegative real entries has a non-negative real eigenvalue. If moreover $M$ has an eigenvector $v$ with strictly positive entries, then the eigenvalue of $v$ is the largest non-negative real eigenvalue and is the spectral radius of $M$.
 A: No, here are infinitely many counter-examples for $I = \{1,2\}$.
Let $n,m \in \mathbb{Z}_{\ge 1}$ such that $m | n^2$, $n | 2m^2$ and $n \neq 2m$ (e.g. $n=m=1$).
Take  $d_1 = n(1+\sqrt{2})$, $d_2 = m(3+2\sqrt{2})$. Here are the matrices $M_i = (n_{i,j,k})_{j,k \in I}$:
$$\left(\begin{matrix}0&\frac{n^2}{m}\\m&2n \end{matrix}\right), \ 
\left(\begin{matrix}m&2n\\\frac{2m^2}{n}&5m\end{matrix}\right)$$
Now $\langle 1 \rangle = \langle 2 \rangle = I$, but $\mathbb{Z}[d_1] = \mathbb{Z}[n \sqrt{2}] \neq \mathbb{Z}[2m \sqrt{2}] = \mathbb{Z}[d_2]$, because $n \neq 2m$.

Here is a counter-example which is also a fusion ring (the initial motivation), with  $I = \{1,2,3\}$.
Take $d_1 = 1$, $d_2 = 3+2\sqrt{2}$, $d_3 = 4+3\sqrt{2}$. Here are the matrices $(M_i)$:
$$\left(\begin{matrix}1&0&0\\0&1&0\\0&0&1 \end{matrix}\right), \ 
\left(\begin{matrix}0&1&0\\1&0&4\\0&4&3 \end{matrix}\right), \
\left(\begin{matrix}0&0&1\\0&4&3\\1&3&6 \end{matrix}\right)$$
Now $\langle 2 \rangle = \langle 3 \rangle = I$, but $\mathbb{Z}[d_2] = \mathbb{Z}[2\sqrt{2}] \neq \mathbb{Z}[3\sqrt{2}] = \mathbb{Z}[d_3]$.
For the notion of fusion ring, see Definition 3.1.7 in the following reference:
P. Etingof, S. Gelaki, D. Nikshych, V. Ostrik; Tensor categories. Mathematical Surveys and Monographs (2015) 205.

Conclusion: We should replace $\mathbb{Z}[d_i] = \mathbb{Z}[d_j]$ by $\mathbb{Q}(d_i) = \mathbb{Q}(d_j)$ in the question.
