Trying to study isomorphism classes of certain commutative Artinian $\mathbb{C}$-algebras I was lead to the following problem about matrices.


Suppose you have a (non-zero) nilpotent matrix $A\in M\_n(\mathbb{C})$. Think of matrix algebras over the Artin algebra $\mathbb{C}[A]$. To be more specific, if another (linearly independent with $A$) nilpotent matrix $B\in M\_n(\mathbb{C})$ which *commutes* with $A$ is given, I want to understand $\mathbb{C}[A]$-isomorphism classes of the subalgebra $\mathbb{C}[A,B]\subset M\_n(\mathbb{C})$ of polynomial expressions in $A,\,B$. Clearly if $M\in GL\_n(\mathbb{C})$ *commutes* with $A$, the conjugated algebra $M(\mathbb{C}[A,B])M^{-1}=\mathbb{C}[A,MBM^{-1}]$ is $\mathbb{C}[A]$-isomorphic to the original one. Here's my question:


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If $\mathbb{C}[A,B]\simeq \mathbb{C}[A,B\_1]$ as $\mathbb{C}[A]$-algebras, is it true that the isomorphism is induced by conjugation with a matrix $M\in GL\_n(\mathbb{C})$ which commutes with $A$? (so $B\_1=MBM^{-1}$)


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I've been trying to work out an example to convince myself in the small case $n=4$. Assume $$A=\begin{pmatrix}0 & 0 & 1 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0\end{pmatrix}.$$ In this case the general form of a - nilpotent - matrix $B$ which commutes with $A$ is $$B\_{yxab}=\begin{pmatrix}0 & y & x & a\\
0 & 0 & 0 & b\\
0 & 0 & 0 & x\\
0 & 0 & 0 & 0\end{pmatrix},$$ where $x,y,a,b$ are arbitrary complex numbers. As we are interested in the algebra $\mathbb{C}[A,B]$ we can simplify $B\_{yxab}$ changing it with $B\_{yxab}-xA-aA^2$ to get a new $$B\_{yb}=\begin{pmatrix}0 & y & 0 & 0\\
0 & 0 & 0 & b\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\end{pmatrix}.$$ Of course (almost all) the $B\_{yb}$'s are conjugated as we can check looking at Jordan forms. However, the conjugation matrices $$\begin{pmatrix}by & 0 & 0 & 0\\
0 & b & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0\end{pmatrix}$$ do not commute with $A$ so the algebras are not $\mathbb{C}[A]$-isomorphic...


My question is related to [Conjugation between commutative subalgebras of a matrix algebra?](https://mathoverflow.net/questions/84591/conjugation-between-commutative-subalgebras-of-a-matrix-algebra), where one of the answers give a counter-example. However it seems that here I have more hypotheses (and more hope...)


Anyway, even if the answer for my question is NO I would appreciate any hint/reference to understand the previously mentioned isomorphism classes. Thanks.



Commuting nilpotent matrices and conjugation isomorphisms
Decided to turn my comment into an answer since it seems to be independent of the ambiguity noticed by @LSpice.


Take
$$
A=
\begin{pmatrix}
0&1&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0
\end{pmatrix},
$$
$$
B\_1=
\begin{pmatrix}
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&1&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0
\end{pmatrix}
$$
and
$$
B\_2=
\begin{pmatrix}
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&1&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&1\\
0&0&0&0&0&0
\end{pmatrix}
$$
Then there are isomorphisms $\mathbb C[A,B\_i]\cong\mathbb C[x,y]/(x^2,xy,y^2)$ carrying $A$ to $x$ and $B\_i$ to $y$, $i=1,2$. But $B\_1$ is not conjugate to $B\_2$ (they both are in Jordan normal forms, which have different number of nonzero blocks).


