Assume a homomorphism of groups gives a full and faithful functor on reps. Was it surjective? Let $\phi: H \to G$ be a finite group homomorphism. Then there is a functor on representations $\operatorname{Rep}(\phi): \operatorname{Rep}(G) \to \operatorname{Rep}(H)$ given by precomposition with $\phi$.
Assume $\operatorname{Rep}(\phi)$ is fully faithful. Is $\phi$ surjective?
 A: Let me call your functor $\phi^{\ast}$. It has a left adjoint (induction) which I'll call $\phi_{\ast}$. You want to know when the natural map
$$\text{Hom}(V, W) \to \text{Hom}(\phi^{\ast} V, \phi^{\ast} W)$$
is an isomorphism. Applying the adjunction, you equivalently want to know when the natural map
$$\text{Hom}(V, W) \to \text{Hom}(\phi_{\ast} \phi^{\ast} V, W)$$
is an isomorphism. By the Yoneda lemma this is equivalent to asking when the natural map
$$\phi_{\ast} \phi^{\ast} V \to V$$
(the counit of the adjunction) is an isomorphism. But $\phi_{\ast} \phi^{\ast}$ multiplies the dimension of a representation by the index $[G : H]$, so we conclude that this is only possible if $[G : H] = 1$.
A: We may replace $H$ by $\phi(H)$ and thus assume $H$ is just a subgroup of $G$, since an $H$-equivariant map between representations in the image of $\operatorname{Rep}(\phi)$ is the same thing as a $\phi(H)$-equivariant map between them.  Now consider the regular representation $V=k[G]$ of $G$.  Note that a $G$-equivariant endomorphism $T:V\to V$ is determined by $T(1)$, since $1$ generates $V$ as a $k[G]$-module.  But as an $H$-representation, $V$ splits as a direct sum of copies of the regular representation $k[H]$, with one copy for each coset of $H$ in $G$.  On each direct summand, we can define an $H$-equivariant endomorphism by right-multiplication by any element of $k[H]$.  In particular, if $T:V\to V$ is $H$-equivariant, $T(1)$ only determines $T$ on the summand corresponding to the coset of $1$.  Thus if $H$ is not all of $G$ (and so there is more than one coset), there are $H$-equivariant endomorphisms of $V$ that are not $G$-equivariant.
