Jordan Decomposition for a certain matrix

This is the form of matrices I am concerned about recently,

Denote by $A=(a_{ij})_{e \times e}$, where $a_{ij}$ is block matrix of dimension $n_i \times n_j$. Now given $n, 0 \leq n \leq e-1$. Let $a_{ij}=0$, except for $j-i=n$ mod $e$. For example, $e=3,n=2$, then

$A= \left( \begin{matrix} 0 & 0 & a_{13} \\ a_{21} & 0 & 0 \\ 0 & a_{32} & 0 \\ \end{matrix} \right)$

My question is, what kind of Jordan decomposition can we get here, especially for its nilpotent part? For another example, $e=2, n=1$ case, we cannot directly use the construction way, because the characteristic polynomial for $A= \left( \begin{matrix} 0 & a_{12} \\ a_{21} & 0 \\ \end{matrix} \right)$ is $x^{n_1-n_2} \times |xI_{n_2}-a_{21}a_{12}|$.

Since there is no further restriction on $A$, I think the answer should be general.

Even the conjugacy classes with respect of block diagonal matrices $\text{diag}\,(a_{11},\ldots,a_{ee})$ are known. More precisely:
The matrix $A$ can be regarded as a cycle of homomorphisms $$k^{n_1}\to k^{n_2}\to\ldots\to k^{n_e}\to k^{n_1}.$$ So it is a representation of the cyclic quiver of type $A_{e-1}^{(1)}$ whose representation theory is well known.
The indecomposable nilpotent blocks look as follows: pick $i$ between $1$ and $e$ and a nilpotence degree $N$. Then a basis of the vector space is $v_1,\ldots, v_N$ and $A$ acts as $Av_\nu=v_{\nu+1}$ and $Av_N=0$. Moreover $k^{n_j}$ is spanned by all $v_\nu$ with $j\equiv \nu+i-1 \text{ mod }e$ (so the $v_\nu$ circle around with period $e$ and starting point $i$). From this it is a purely combinatorial problem to find all possible Jordan forms. For example, a single Jordan block is only possible if the dimension vectors are of the form $(a+1,\ldots,a+1,a,\ldots,a)$ or a cyclic permutation thereof.
You can find more, e.g., in Kempken's thesis "Eine Darstellung$\ldots$" MR0666395 (84a:14022) of 1982.