If you are familiar with the algebra, just skip the following brief introduction, that is fine.
In the study of Bose-Mesner algebra. We know given that the commutative association scheme $\mathfrak{X}:=(X,\{R_{i}\}_{0\leq i\leq d}).$ Its Bose-Mesner algebra $\mathfrak{A}$ has two natural basis, that is, the adjacency matrices $A_{0},A_{1},...,A_{d}$ and the idempotent matrices $E_{0},E_{1},...,E_{d}.$ In the latter case, we denote $\mathfrak{A}^{\circ}$ instead of $\mathfrak{A}$ since it is the algebra with respect to the Hadamard product. For fix $0\leq i\leq d,$ we have
\begin{align} A_{i}=\sum_{j=0}^{d}P_{i}(j)E_{j}\quad\text{and}\quad E_{i}=\frac{1}{|X|}\sum_{j=0}^{d}Q_{i}(j)A_{j} \end{align} These two base change matrices $P=(P_{j}(i))_{0\leq i,j\leq d}$ and $Q=(Q_{j}(i))_{0\leq i,j\leq d}$ are called the first eigenmatrix and the second eigenmatrix, respectively. On the other hand, from the representation theory, there are left regular representations $\rho:\mathfrak{A}\rightarrow M_{d+1}(\mathbb{C})$ and $\rho^{*}:\mathfrak{A}^{\circ}\rightarrow M_{d+1}(\mathbb{C}).$ Moreover, they are isomorphisms of algebras $\rho:\mathfrak{A}\rightarrow\mathfrak{B}\subset M_{d+1}(\mathbb{C})$ and $\rho^{*}:\mathfrak{A}^{\circ}\rightarrow\mathfrak{B}^{*}\subset M_{d+1}(\mathbb{C}),$ where $M_{d+1}(\mathbb{C})$ is the full matrix algebra of degree $d+1$ over $\mathbb{C}$ and $\mathfrak{B}$ is so called Intersection Algebra and $\mathfrak{B}^{*}$ as its dual.
Sorry for long introduction, now, I just think about two questions as follows.
Q1: Are the representations $\rho$ and $\rho^{*}$ similar ? That is, does there exist a non-singular matrix $U$ such that $\rho=U\rho^{*}U^{-1}.$ I think the answer should be positive, but I am not very sure.
Q2: Do we have a natural morphism or map from $\mathfrak{B}$ to $\mathfrak{B}^{*}$? If so, is this a commutative diagram ? In the picture, we have
$\require{AMScd}$ \begin{CD} \mathfrak{A} @>{\rho:A_{i}\mapsto}B_{i}>> \mathfrak{B}\\ @V P VV \\ \mathfrak{A}^{\circ} @>{\rho^{*}:|X|E_{i}\mapsto}B_{i}^{*}>> \mathfrak{B}^{*} \end{CD}
where $\mathfrak{B}=\langle B_{0},B_{1}...,B_{d}\rangle$ and $\mathfrak{B}^{*}=\langle B_{0}^{*},B_{1}^{*}...,B_{d}^{*}\rangle.$
Any comment or advice I will be grateful. Thanks for patient reading.