= Solution
A <group representation> on a <vector space> $V$ over a <field> $K$ is a <group homomorphism> $R:G\to\operatorname{GL}(V)$. In particular, $R(gh)=R(g)R(h)$, $R(e)=I$ and $R(g^{-1})=R(g)^{-1}$. Choosing a <basis> identifies it with a <matrix representation>.
Two <group representations> $R$ on $V$ and $S$ on $W$ are <equivalent representations> when an invertible <linear map> $T:V\to W$ satisfies $TR(g)=S(g)T$ for every $g$. Thus $S(g)=TR(g)T^{-1}$; they differ only by a choice of <basis>.
A nonzero <group representation> is a <reducible representation> if there is a <vector subspace> $U$ with $0\ne U\ne V$ and $R(g)U\subseteq U$ for every $g$. It is an <irreducible representation> if it has no such <invariant subspace>. Reducibility alone does not assert the existence of an invariant complementary <vector subspace>; that stronger conclusion says that the <representation> is a <semisimple representation>.
The <Schur lemma> states that an <intertwiner> $T$ between <irreducible representations> is either zero or an <isomorphism>. Consequently an <intertwiner> between inequivalent <irreducible representations> is zero. For a finite-dimensional <irreducible representation> over the <complex numbers>, every commuting <endomorphism> is scalar:
$$
\boxed{TR(g)=R(g)T\quad(\forall g)\quad\Longrightarrow\quad T=\lambda I.}
$$
For the first assertion, the <kernel> and <image> of $T$ are <invariant subspaces>, so a nonzero $T$ is injective and surjective. For the scalar conclusion, choose an <eigenvalue> $\lambda$ of $T$. The <intertwiner> $T-\lambda I$ has a nonzero <kernel>, so it must vanish. The finite-dimensional complex hypothesis matters: the scalar conclusion need not hold over the <real numbers>, as a real planar rotation through $\pi/2$ is an <irreducible representation> of the cyclic <group> of order four, while the rotation itself is a commuting nonscalar <endomorphism>.
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