Question in the proof of Clifford Theorem (representation theory) 
Let $A$ be a normal subgroup of $G$. Let $\rho:G\to GL(V)$ be a
irreducible representation. Write $Res^G_A\rho$ as a direct sum of
isotypic representations of $A$;$V=\bigoplus\limits_{i=1}^k W_i $. Since
$A$ is normal subgroup of $G$, $\rho(g)W_i$ is stable under $A$. Since
$W_i$ is isotypic, $\rho(G)W_i$ is also isotypic and conjugate to
$W_i$ so $G$ permutes $W_i$'s. Since $V$ is irreducible representation
of $G$, we know that the action of $G$ on $W_i$'s is transitive and
hence all $W_i$'s are conjugate.

I understand that $\rho(g)W_i$ is stable under A. But I don't understand the next two sentences. Can you please explain what is going on here? I would really appreciate any help.
 A: Any group element $g \in G$ defines an automorphism on the normal subgroup $A$ by conjugation: $c_g(a) = g^{-1} a g$. For any representation $\pi \colon A \to \operatorname{GL}(W)$ of $A$, we can pull the representation back along this automorphism to get another representation $\pi \circ c_g$ of $A$, namely the representation where we have $a \cdot w = \pi(g^{-1} a g) w$. If $\pi$ is irreducible then so is $\pi \circ c_g$, so precomposing by $c_g$ permutes the isomorphism classes of irreps of $A$. Accordingly if $\pi$ is isotypic then so is $\pi \circ c_g$.
Now suppose we have a representation $\rho \colon G \to \operatorname{GL}(V)$. We can treat $\rho$ as a representation of $A$, yielding new representations $\rho \circ c_g$ as before, but now we can factorise $\rho(g^{-1} a g) = \rho(g^{-1}) \rho(a) \rho(g)$ because $\rho$ is defined on $G$ rather than on $A$. The normality of $A$ implies that for any $A$-subrepresentation $W \subseteq V$, the subspace $\rho(g) W$ is again $A$-stable, and hence $(\rho(g) W, \rho)$ is another $A$-subrepresentation of $V$. In fact, left multiplication by $\rho(g)$ gives an isomorphism $(W, \rho \circ c_g) \to (\rho(g) W, \rho)$ which is $A$-equivariant:
$$ \rho(g)(\rho \circ c_g)(a) w = \rho(g g^{-1} a g) w = \rho(a) \rho(g) w.$$
So the conjugated representation $(W, \rho \circ c_g)$ of $A$ is isomorphic to the subrepresentation $(\rho(g) W, \rho)$. So something that used to be quite abstract (changing the action on a space by precomposition with an automorphism) becomes quite concrete (another subrepresentation in the same ambient space).
Now specialising to your question: by "$\rho(g) W_i$ is conjugate to $W_i$" they mean in the sense above: $(\rho(g) W_i, \rho)$ is isomorphic to the conjugated representation $(W_i, \rho \circ c_g)$. Since $W_i$ is isotypic it is of the form $W_i \cong X^{\oplus k}$ for some particular irrep $X$ of $A$. Since conjugation permutes the irreps of $A$, it must permute the isotypic components of $V$, hence $\rho(g) W_i = W_j$ for some $j$ depending on $i$ and $g$.
Lastly, for each $W_i$ and $W_j$ there must exist a $g \in G$ such that $\rho(g) W_i = W_j$, since if this were not the case $V$ would not be irreducible as a representation of $G$, as $\sum_{g \in G} \rho(g) W_i$ would be a proper $G$-stable subspace.
