 Solution

Work over the complex numbers. By Maschke's theorem, restriction to the normal subgroup $A$ is a direct sum of its isotypic components. For an irreducible $A$-type $W$, denote its isotypic component in $V$ by $V_W$. Normality makes $G$ permute these components, by conjugating the $A$-types. Any union of orbits spans a $G$-invariant subspace, so irreducibility of $V$ means that all occurring components form a single orbit.

If there is only one component, restriction is isotypic, as claimed. Otherwise take one component $V_W$ and its stabilizer $H=\{g:g{V}_W=V_W\}$, a proper subgroup. This is an irreducible representation of $H$: if $0\ne U⊊V_W$ were $H$-invariant, the direct sum of its translates over coset representatives for $G/H$ would be a nonzero proper $G$-invariant subspace of $V$. It is direct because the translates lie in distinct isotypic components, and proper because $\dim U<\dim V_W$.

The natural map from the induced representation sends $g\otimes v$ to $gv$. It identifies the direct-sum coset decomposition with all isotypic components of $V$, so is bijective and $G$-equivariant. ConsequentlyThis proves the dichotomy directly, including the irreducibility of the representation being induced.