Commuting nilpotent matrices and conjugation isomorphisms 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?, 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.
 A: Assuming that you are asking for an isomorphism taking $B$ to $B$':
No. Take $A$ to be the $n\times n$ Jordan block with eigenvalue zero. The centralizer of $A$ is exactly $\mathbb C[A]$. Hence for any $B,B' \in \mathbb C[A]$, we have $\mathbb C[A,B] = \mathbb C[A] = \mathbb C[A,B']$. However, $B$ and $B'$ need not be conjugate: take $B = A^2, B' = A^3$. As long as $n \geq 4$, $\{1,A,A^2,A^3\}$ is linearly independent, and none are conjugate (since their kernels have different dimensions).
A: 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).
