$\text{dim}(\text{Hom}^G(V,W))=1$ implies that isomorphism of representations Let $(\pi, V)$ and $(\rho, W)$ be two irreducible representations.
I am trying to prove that $\text{dim}(\text{Hom}^G(V,W))=1$ if and only if $(\pi, V)$ and $(\rho, W)$ are isomorphic.
I have shown here that when $(\pi, V)$ and $(\rho, W)$ are isomorphic it follows that $\text{dim}(\text{Hom}^G(V,W))=1$ but I do not see how to go about doing the other direction.
I feel that if $\text{dim}(\text{Hom}^G(V,W))=1$ then $(\text{Hom}^G(V,W))$ consists of $1\times 1$ matrices? Then they would have to be invertible? But this does not seem correct.
 A: Let's take a closer look in Schur's Lemma. In its most common form, it says that:

Let $(\pi,V)$ and $(\rho,W)$ be two irreductible representations (over $\mathbb{C}$) of $G$. Then:

*

*If $(\pi,V)$ and $(\rho,W)$ are not isomorphic, then there are no nontrivial equivariant maps between them.

*If $V=W$ and $f$ is an equivariant map between them, then $f=\lambda\:\mathbb{id}$, for some $\lambda\in\mathbb{C}$.


For the first point, it says that $\hom^G(V,W)$ consists of only the trivial map. In other words, $\hom^G(V,W)$ is the trivial vector space and hence it is $0$-dimensional.
For the second point, we have to do just a little more work. The theorem says that $\hom^G(V,V)$ consists of all the maps on the form $\lambda\:\text{id}$. Thus $\{\text{id}\}$ is a basis for $\hom^G(V,V)$ and it is $1$-dimensional. But what about $\hom^G(V,W)$, when $(\pi,V)$ and $(\rho,W)$ are isomorphic? Well, if $F:V\to W$ is an equivariant isomorphism, then
\begin{align*}
\hom^G(V,V) &\to \hom^G(V,W)\\
 h &\mapsto F\circ h
\end{align*}
is an isomorphism between $\hom^G(V,V)$ and $\hom^G(V,W)$. We conclude that $\hom^G(V,W)$ is also $1$-dimensional.
This shows that Schur's Lemma is precisely the statement below.

Let $(\pi,V)$ and $(\rho,W)$ be two irreductible representations (over $\mathbb{C}$) of $G$. Then, $\hom^G(V,W)$ is $1$-dimensional if and only if $(\pi,V)$ and $(\rho,W)$ are isomorphic. Else, if they are not isomorphic, $\hom^G(V,W)$ is $0$-dimensional.

A: This answer is based of Jyrki's comments. 
Suppose that $\text{dim}(\text{Hom}^G(V,W))=1$. Then there is a non zero map $\phi\in \text{Hom}^G(V,W)$. 
It can be shown that if $\phi$ is non zero, then it must be an isomorphism. 
To do this show that $\ker(\phi)$ and $\text{Im}(\phi)$ are invariant subspaces (This is shown in the linked question). Next use that the representations are irreducible. Since $\phi$ is nonzero $\text{Im}(\phi)=W$. This gives that $\ker(\phi)\neq V$ and hence $\phi$ is an isomorphism and we are done.
