Let be $C(u)$ the set of commuting endomorphisms with $u$ over a complex-valued vector space of finite dimension $n \in \mathbb{N}^{*}$.
I call $v$ a cyclic endomorphism iff all its eigenspaces are one dimensional vector space.
I'd like to know if $C(u)$ contains at least always a cyclic endomorphism.
My proof strategy is to distinguish multiple cases:
(1) if $u$ is diagonalizable: I can build such an invertible cyclic endomorphism by considering the eigenspaces
(2) if $u$ is nilpotent: that's precisely my issue, I'm trying to mess with Jordan reduction, but failed to do so.
(3) general case: I write $u = d + n$, such that $d$ is diagonalizable and $n$ nilpotent and $dn = nd$, I get an invertible cyclic endomorphism $v$ which commutes with $d$, but that's not enough to get $vu = uv$.
Your conjecture is true.
Let $A\in M_n(K)$ where $K$ is an algebraically closed field.
Then the Jordan form of $A$ is as follows
$A=Pdiag(\lambda_1 I_{k_1}+J_{k_1},\cdots,\lambda_p I_{k_p}+J_{k_p})P^{-1}$ where $J_r$ is the nilpotent Jordan block of dimension $r$, $k_1+\cdots+k_p=n$ and the $\lambda_i \in K$ are eigenvalues of $A$ that are NOT NECESSARILY DISTINCT.
Let $\mu_1,\cdots,\mu_p$ be $p$ DISTINCT elements of $K$.
Let $B=Pdiag(\mu_1 I_{k_1}+J_{k_1},\cdots,\mu_p I_{k_p}+J_{k_p})P^{-1}$; then $B\in C(A)$ and is cyclic.
REMARK. On a general field $K$, a cyclic matrix is s.t. its characteristic and minimal polynomials are the same. I don't know (I did not think about) if the previous result holds in this general case.