Morphisms of two fully reducible representations of a group Consider fully reducible representations $\,A(g)\,$ and $\,A^{\,\prime}(g)\,$ of a group $\,G\,$ in vector spaces $\,\mathbb{V}\,$ and $\,{\mathbb{V}}^{\,\prime}\,$, respectively. Let them be intertwined:
 $$
 M~A(g)~=~A^{\,\prime}(g)~M~~.
 $$
 For brevity, I shall denote the kernel $\,\mbox{Ker}\,M\,$ simply as $\,$Ker, the image $\,\mbox{Im}\,M\,$ as $\,$Im$\,$. Being invariant subspaces they support subrepresentations:
 $$
 B(g)\,v~\equiv~A(g)\,v\Big{|}_{v\,\in\,\rm{Ker}}\quad,\qquad B^{\,\prime}(g)\,v^{\,\prime}~\equiv~A^{\,\prime}(g)\,v^{\,\prime}\Big{|}_{v^{\,\prime}\,\in\,\rm{Im}}\;\;.
 $$
 As $\,A\,$ is fully reducible, any of its subrepresentations has a complementary subrepresentation.
 E.g., for $\,B\,$ acting in $\,$Ker$\,$, its complementary $\,B^{\,\perp}\,$ in $\,{\mbox{Ker}}^{\perp}\,$ is
 $$
 B^{\perp}(g)\,v\,\equiv\,A(g)\,v\Big{|}_{v\,\in\,{\rm{Ker}}^{\perp}}\;\;.
 $$
It is then easy to demonstrate that the same $\,M\,$ intertwines $\,B^{\perp}\,$ and $\,B^{\,\prime}\,$,
 i.e. $\,M\,B^{\perp}\,=\,B^{\,\prime}\,M\,$. Moreover, if we postulate $\,B^{\,\prime}\,$ to be irreducible, the representations $\,B^{\perp}\,$ and $\,B^{\,\prime}\,$ become equivalent, by Schur's Lemma:
 $$
 B^{\perp}\,\simeq\,B^{\,\prime}\;\;.
 $$
 The inverse is true too: if $\,B^{\perp}\,\simeq\,B^{\,\prime}\,$, there exists a morphism $\,M\,$ intertwining $\,A\,$ and $\,A^{\,\prime}\,$.
To conclude, fully reducible representations $\,A\,$ and $\,A^{\,\prime}\,$ intertwine if and only if they have equivalent subrepresentations.
 $$
 ~~
 $$
 QUESTION 1:  $~~~$In the case of an irreducible $\,A\,$, prove that the multiplicity of $\,A\,$ in $\,A^{\,\prime}\,$ is equal to the dimensionality of the space
 $\,[A\,,\,A^{\,\prime}]\,$ of all such intertwiners $\,M\,$.
 $$
 ~~
 $$
 QUESTION 2:  $~~~$If $\,$dim$\,[A\,,\,A^{\,\prime}]\,=\,\,\infty\,$, would it be right to say that the representations $\,A\,$ and $\,A^{\,\prime}\,$ are equivalent, and their spaces are isomorphic?
 A: For Question $1$, I'll assume that we're in a situation where $A$ satisfies the second part of Schur's Lemma: every map of representations $\mathbb{V}\to\mathbb{V}$ is multiplication by some scalar $\lambda\in\mathbb{C}$. Or in the language of the question, if $N$ is such that $N~A(g)=A'(g)~N$ for all group elements $g$, then $N$ is a scalar multiple of the identity.
Suppose that $A$ has multiplicity $m$ in $A'$. So we can decompose $\mathbb{V}'$ as a direct sum
$$\mathbb{V}'=\mathbb{W}_1\oplus\dots\oplus \mathbb{W}_m\oplus \mathbb{U}$$
of invariant subspaces such that the representations on $\mathbb{W}_1,\dots,\mathbb{W}_m$ are all equivalent to $A$, and where the representation on $\mathbb{U}$ has no subrepresentation equivalent to $A$.
So there are invertible morphisms ("intertwiners") $\varphi_i:\mathbb{V}\to\mathbb{W}_i$ for $i=1,\dots,m$ such that every morphism $\mathbb{V}\to\mathbb{W}_i$ is a scalar multiple of $\varphi_i$. And there are no nonzero morphisms $\mathbb{V}\to\mathbb{U}$.
Now define morphisms $\varphi'_i:\mathbb{V}\to\mathbb{W}_1\oplus\dots\oplus \mathbb{W}_m\oplus \mathbb{U}=\mathbb{V}'$ by
$\varphi'_i(v)=(0,\dots,0,\varphi_i(v),0,\dots,0)$, where the only nonzero component is $\varphi_i(v)\in\mathbb{W}_i$.
I claim that $\varphi'_1,\dots,\varphi'_m$ form a basis of the space of all morphisms $\mathbb{V}\to\mathbb{V}$, and so $m$ is the dimension of this space.
It is straightforward to check that they are linearly independent, so we just need to prove that every morphism is a linear combination of them.
Let $\theta:\mathbb{V}\to \mathbb{V}'$ be a morphism, and write 
$$\theta(v)=(\theta_1(v),\dots,\theta_m(v),\theta_U(v)).$$
Then it straightforward to check that $\theta_i:\mathbb{V}\to\mathbb{W}_i$ is a morphism, and so $\theta_i=\lambda_i\varphi_i$ is a scalar multiple of $\varphi_i$. Also $\theta_U:\mathbb{V}\to\mathbb{U}$ is a morphism, and is therefore zero.
So 
$$\theta=\lambda_i\varphi'_1+\dots+\lambda_m\varphi'_m$$
is a linear combination of $\varphi'_1,\dots,\varphi'_m$ as required.
