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?

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 $[G : H] = 1$.