(Assumptions): Let $G$ be a group and $\rho_1:G\to Aut(V)$, $\rho_2:G\to Aut(W)$ be two irreducible representations of $G$ in finite-dimensional vector spaces $V, W$ over an algebraically closed field $\mathbb{K}$. Let us note the Schur's lemma and its modified version:
(Schur's lemma): If $V$ and $W$ are irreducible representations of a group and $f:V\to W$ is an intertwining map, then either $f \equiv 0$ or $f$ is an isomorphism
(Schur's lemma: Modified) If the assumptions regarding $G, V$ and $\mathbb{K}$ hold, then any intertwining map $f:V\to V$ is of the form $f = c \cdot \mathrm{id}_V, c \in \mathbb{K}$
(Question): My source materials claims that the dimension of the space of intertwining maps from $V$ to $W$ is one, if $V$ is isomorphic to $W$, and zero otherwise. The case for zero follows easily from the Schur's lemma above. But I'm struggling to understand the case when $V\cong W$: My source states that the claim follows immediately from the modified Schur's lemma. But how can one argue about it algebraically? Conceptually it does make sense, as the representations $\rho_1, \rho_2$ have the same structure under isomorphism. But what exactly would be our intertwining map $f:V\to W \cong V$?
Best Answer
It's important to understand what is meant by "if $V$ is isomorphic to $W$" - it means as representations, i.e. that there exists at least one intertwining map $f:V\to W$. If this is the case, it's not too difficult to show using modified Schur that any other intertwiner is proportional to the first (hint: just take any other intertwiner $f':V\to W$, and consider $f^{-1}\circ f'$).