Let $C,D$ be a categories, $G$ a group and $g:G\to\text{Aut}( C)$, $h:G\to\text{Aut}(D)$ group homomorphisms (into the group of classes of autoequivalences). Further assume $g$ to be injective and let $F:C\to D$ be an equivalence such that the following diagram commutes for all $a\in G$:
$$F\circ g(a)=h(a)\circ F.$$
Is it true that $h$ is injective as well?
Assume $h(a) \simeq \mathrm{id}_D$. Then $h(a)\circ F \simeq F \simeq F\circ g(a)$.
Compose with a quasi-inverse $H$ of $F$ to get $g(a)\simeq \mathrm{id}_C$.
Thus if your "homomorphisms" are weak homomorphisms in the sense that $g(ab)\simeq g(a)g(b)$, and $g$ is "strongly injective" in the sense that $g(a)\simeq g(b) \implies a=b$, then this implies $a=$ the neutral element; so indeed $h$ is strongly injective as well.
However, if you stick with "normal" injectivity (i.e. $g(a)=g(b) \implies a=b$) then you can't prove that $h$ is injective too.
Consider indeed the trivial connected groupoïd with two objects $C$ and $D$ the trivial group. Then $Aut(C)\simeq \mathbb{Z/2Z}$, $Aut(D)\simeq 1$.
Put $G= \mathbb{Z/2Z}$, $g$ the unique isomorphism with $Aut(C)$, and $h$ the unique morphism to $Aut(D)$. Let $F:C\to D$ be the equivalence. Then $F\circ g(a) = h(a)\circ F$ for $a\in G$, but clearly $h$ is not injective.
So from "strong injectivity" of $g$ you can get "strong injectivity" of $h$, but you can't get injectivity of $h$ from injectivity of $g$.