Fix a real number $n$. By a "matrix", I mean an $n \times n$ real matrix. Now let $A$ denote a matrix. Is it true that for all traceless matrices $T$, there exists a traceless matrix $T'$ such that $AT = T'A$?
(This is certainly true whenever $A$ is invertible, by taking $T' = ATA^{-1}$.)
Let's take the situation where $A$ is not invertible.
It suffices to note that there exists some matrix $R$ such that $RA = AT$, and that there exists some matrix $S$ with non-zero trace such that $SA = 0$. It follows that there exists some $\alpha \in \Bbb R$ such that the matrix $T' = R + \alpha S$ is traceless. Moreover, this matrix satisfies $$ AT = T'A $$ as desired.