An epigroup is a semigroup $S$ in which every element has a (positive) power that lies in a subgroup of $S$. (The subgroup may depend on the element). Note that if $x^n\in G$, where $G$ is a subgroup of $S$, then $x^m\in G$ for all $m\ge n$.
It is stated on the wiki article on epigroups that:
The semigroup of all matrices over a division ring is an epigroup.
It seems to be taken directly from their source, The Concise Handbook of Algebra (Shevrin 2002), though I think what they mean must be the following (otherwise we can't talk of a semigroup at all):
The semigroup of all square matrices of a given size over a division ring is an epigroup.
Why is this true? I believe I have stumbled my way to a proof that works when the matrices are over an algebraically closed field, since this lets us use Jordan normal forms, but I don't know if it extends to even the case of general fields. Going from general fields to skew fields seems even harder, since this leaves the realm of standard linear algebra with which I'm familiar. I do suspect that there are nicer, higher-level approaches that don't use a concrete structure of the matrices like Jordan matrices. I would appreciate a more abstract proof.
In my proof, we assume without loss of generality that $X$ is a Jordan matrix. This of course requires the existence of a Jordan normal form for any matrix, which is why I think I need an algebraically closed field. Let $n$ be the size of the largest Jordan block with eigenvalue $0$. Then the nilpotent Jordan blocks of $X$ have vanished in $X^n$, so we get a block matrix of the form $$ X^n = \begin{pmatrix}J_{\lambda_1}^n\\&\ddots\\&&J_{\lambda_k}^n\\&&&0\end{pmatrix}, $$ where $J_{\lambda_j}$ is a Jordan block of eigenvalue $\lambda_j\ne0$. Let $E$ to be the diagonal matrix with $1$'s at all the non-zero diagonal places of $X^n$. Then $E$ is an idempotent, and $X^nE=X^n=EX^n$. Furthermore, Jordan blocks are invertible (by the formula in this answer), so if we take $$ Y=\begin{pmatrix}J_{\lambda_1}^{-n}\\&\ddots\\&&J_{\lambda_k}^{-n}\\&&&0\end{pmatrix}, $$ then $YX^n=E=X^nY$, which shows that $X^n$ is included in a subgroup with identity $E$. Translating the results to general (non-Jordan) $X$ is easy from here.
Suppose $V$ is a finite-dimensional vector space and $X:V\to V$ is linear. Let $V_n$ be the image of $X^n$. Since the $V_n$ are a decreasing sequence of subspaces of $V$, they must eventually stabilize (their dimension can only go down finitely many times). So there is some $W\subseteq V$ such that $V_n=W$ for all sufficiently large $n$. This implies that for $n$ large, $X^n$ has image $W$ and also maps $W$ surjectively onto itself. Let $K=\ker(X^n)$ for such a large $n$. Since $X^n$ maps $W$ surjectively to itself, $V=W\oplus K$.
Now let $G$ be the set of endomorphisms of $V$ that have kernel $K$ and image $W$. Note that restriction to $W$ gives a composition-preserving bijection between $G$ and $GL(W)$, so $G$ is a subgroup of $\operatorname{End}(V)$ that contains $X^n$.