A professor of mine told me that this is true, but he doesn't remember what the proof was or where to find it, and I haven't been able to find a source for it yet. As such I am looking for one here.

In the theorem as stated, $\mathbb{F}$ is any field and $T_n(\mathbb{F})$ denotes the algebra of upper triangular $n\times n$ matrices over $\mathbb{F}$.

Theorem: Let $A,B\in T_n(\mathbb{F})$ be such that for all $X\in T_n(\mathbb{F})$, $$AX=XA\implies BX=XB$$ Then $B=p(A)$ for some $p\in \mathbb{F}[t]$.

Does anyone know of a source for this result? I have searched Google, MSE, MO, and the like to no avail.

If we replace $T_n(\mathbb{F})$ by $M_n(\mathbb{F})$, the question is answered in this paper. Unfortunately, the argument doesn't seem to translate directly, as I can't find a way to force the $M_i$ maps to be upper-triangular.

Also, I have already asked this question here on MSE. As the question is for an undergraduate research project, it felt appropriate to ask it here as well.

Thanks for any help!

Edit on 9 July, 2018: It's probably worth mentioning that the following theorem is false, so an appeal to Jordan form won't work (at least, not as easily as we'd hope it would).

Fake Theorem: If $A\in T_n(\mathbb{F})$, then there exists an invertible $T\in T_n(\mathbb{F})$ and a permutation matrix $P$ such that $P^{-1}T^{-1} ATP$ is in Jordan form.

An explicit counterexample is $$A=\left[\begin{array}{cccc} 0&1&0&0\\ &0&0&1\\ & &0&1\\ & & &0\end{array}\right]$$ and a more detailed demolishing of this theorem is given here, where the authors prove that if $n\geq 12$ and $\mathbb{F}$ is infinite, then there are infinite sets of nilpotent matrices in $T_n(\mathbb{F})$, none of which are conjugate (in $T_n(\mathbb{F})$) to any of the others.

I mention this because I thought it was true for longer than I'd like to admit, and a few other people I've talked to thought it was true as well until told otherwise.