# Why does Schur's Lemma imply $\dim \hom(V_1,V_2)^G=1$ if $V_1\cong V_2$?

Schur's Lemma says:

Let $$\rho_1:G\to GL(V_1)$$ and $$\rho_2:G\to GL(V_2)$$ be two irreductible representations over $$\mathbb{C}$$ of a group $$G$$. If $$f:V_1\to V_2$$ is an equivariant linear transformation, then:

1) Either $$f$$ is an isomorphism or $$f=0$$ and

2) If $$V_1=V_2$$ and $$\rho_1=\rho_2$$, then $$f=\lambda\text{ id}$$ for some $$\lambda\in\mathbb{C}$$.

Surely, this implies $$\dim \hom(V,V)^G=1$$. What I don't understand is why it implies that $$\dim \hom(V_1,V_2)^G=1$$ if $$V_1\cong V_2$$.

If $$V_1$$ and $$V_2$$ are isomorphic but not equal, we cannot prove that $$f$$ has an eigenvalue as it doesn't even make sense. So it seems to me that Schur's Lemma does not apply here.

If $$(V,\rho_1)$$ is $$G$$-isomorphic to $$(W,\rho_2)$$ then $$\hom_G(V,W)$$ is linearly isomorphic to $$\hom_G(V,V)$$. To prove this you only need to know that $$\hom_G(V,-)$$ is a functor. Functors map isomorphisms to isomorphisms.

For a direct proof: if $$f:V\to W$$ is a $$G$$-isomorphism then the induced map

$$\hom_G(f):\hom_G(V,V)\to\hom_G(V,W)$$ $$\hom_G(f)(h)=f\circ h$$

is a linear isomorphism. The inverse is given by $$\hom_G(f^{-1})$$.