Let $\mathcal E \subset M_5(\mathbb R)$ be a subset of matrices defined by: for every $A \in \mathcal E$, $A$ admits the structure \begin{align*} A = \begin{pmatrix} 0 & * & 0 & 0 & * \\ 1 & * & 0 & 0 & * \\ 0 & * & 0 & 0 & * \\ 0 & * & 1 & 0 & *\\ 0 & * & 0 &1 & * \end{pmatrix}, \end{align*} where $*$ means it can assume any real value.
Suppose we have a monic polynomial $p(t) = (t-a)^5$ with $a \in \mathbb R$. I am wondering whether the matrices in $\mathcal E$ such that $\chi_{A}(t) = p(t)$ have uniquely defined Jordan blocks (up to permutation) where $\chi_A(t)$ denotes the characteristic polynomial of $A$. Apparently, if we realize $p(t)$ in block diagonal manner, i.e., \begin{align*} A = \begin{pmatrix} 0 & -a^2 & 0 & 0 & 0 \\ 1 & 2a & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & a^3 \\ 0 & 0 & 1 & 0 & -3a^2\\ 0 & 0 & 0 &1 & 3a \end{pmatrix}, \end{align*} then there are two Jordan blocks, with size $2$ and size $3$. I feel that other realizations should give the same Jordan blocks, but not sure how to formally argue this point.
This is not true. For the $A$ below, we have $A^3\ne0=A^4$: $$ A=\pmatrix{ 0&0&0&0&0\\ 1&0&0&0&0\\ 0&0&0&0&0\\ 0&1&1&0&0\\ 0&1&0&1&0}. $$