Matrices commuting with an irrep up to automorphism Let $R$ be an irreducible representation of a finite group $G$.
Schur's lemma tells us that matrices $M$ that commute with $R$
$$MR(g)=R(g)M~, \forall g\in G$$
are proportional to the identity
$$M=\lambda I~.$$
What can be said about matrices $M$ that commute with $R$ up to an automorphism $\sigma$ of $G$?
$$MR(g)=R(\sigma(g))M~, \forall g\in G$$
 A: Following the comment, such a matrix will exists if and only if $R$ and $R\circ \sigma$ are isomorphic as representations of $G$, and all such matrices intertwining these representations will be scalar multiples, by Schurs lemma. So essentially there is one matrix $M$, up to scaling with this property.
This matrix $M$ is incredibly interesting however, if it exists, and the automorphism $\sigma$ is order $n$, then $M^n$ is a scalar matrix, so we may assume (by scaling) that $M^n=I$. This gives you the data you need to extend the representation to an irrep of the semidirect product group of $G$ and $\langle \sigma \rangle$.
To give an impression of the depth of the phenomena at play here, if the order of $\sigma$ is coprime to $|G|$, and this matrix $M$ exists, the trace of $M$ (normalised so $M^n=I$) divides the dimension of $V$, and this theorem was only proven recently, before which it was described as one of the deepest problems in character theory by Navarro. See this paper: https://academic.oup.com/blms/article-abstract/26/6/513/400882?redirectedFrom=PDF.
