Schur's Lemma: Is the isormorphism between two irreducible spaces unique? Suppose $V_1 \neq V_2$ are two irreducible representations of the finite group G. Then Schur's Lemma says that any G-invariant map between them is either 0 or an Isormorphism. I understand that if $V_1 = V_2 = V$ then any automorphism is a scalar multiple and it is proved by considering eigen spaces.
QHowever if these two vector spaces are not the same then what can I say about the isomorphism between them? Is it unique upto a scalar multiple? How does one prove this since I cannot consider eigen spaces.
 A: I'm going to speak more generally of $A$ modules for an $F$-algebra $A$, $F$ a field. This would apply to the algebra $F[G]$ for use in representation theory.
As Schur's theorem says, $D=\mathrm{End}(V_A)$ is a division ring if $V$ is a simple right $A$ module.
Now, for any particular element $a\in D$ and $f\in F$, $fa$ is indeed another isomorphism of $V$, but in principal there are more than just the $F$ scalar multiplies. In fact, all of the $D$ scalar multiples are isomorphisms as well! (Elements of $d$ multiply on the left of the isomorphism.) You see, $D$ can be larger than $F$, so there might be more things than just the $F$ multiples. So you can say that the isomorphisms are unique up to $D$ multiples, but I don't think that's what you were originally aiming for.
Now, $D$ can't be too much larger: $D$ is a finite dimensional-$F$ algebra.
If your field $F$ happened to be algebraically closed (like $\Bbb C$, for example), then there are no proper finite dimensional extensions of $F$, so then $D=F$. In that case, yes any isomorphism of $V$ you pick is an $F$ multiple of each other isomorphism.
