Equivalence of categories preserve automorphism group Let $C$ and $D$ be two categories related by the equivalence of categories $F \colon C \to D$. Then I want to show that $\mathrm{Aut}(c) \simeq \mathrm{Aut}(F(c))$.
This seems true, but I am having trouble really writing a proof down. My strategy so far has been as follows:
Let $G$ be an “inverse” of $F$. These functors induce group homomorphisms $F^*$ and $G^*$. Using the natural isomorphism with the identity I can show that $\mathrm{Aut}(c)$ is isomorphic to $\mathrm{Aut}(GF(c))$. And so $G^* \circ F^*$ is equal to an isomorphism from $\mathrm{Aut}(c)$ to $\mathrm{Aut}(GF(c))$. From here it shouldn’t be too hard to modify $F^*$ or $G^*$ to get morphisms which are mutually inverse. But I am not succeeding. I feel like I am missing something obvious, any help would be greatly appreciated.
 A: Any equivalence of categories $F: \mathcal{C} \to \mathcal{D}$ is:

*

*full: surjective on Hom-sets,

*faithful: injective on Hom-sets,

*essentially surjective: for every object $D$ in $\mathcal{D}$ there is an object $C$ in $\mathcal{C}$ such that $F(C)$ is isomorphic to $D$.

In fact, some people take the above as a definition for being an equivalence of categories,  because when you assume a big enough version of the axiom of choice you can construct an "inverse" to $F$ (see e.g. nLab, Wikipedia and this question).
In particular this means that $F$ induces a bijection between $\operatorname{Aut}(C)$ and $\operatorname{Aut}(F(C))$ for any object $C$ in $\mathcal{C}$. Since $F$ also induces a group homomorphism between these two groups, it induces a group isomorphism. Note that this only uses $F$ being full and faithful, so the essential surjectivity is no necessary.
If you really want to write down an explicit inverse for the group isomorphism you can go through the proofs of $F$ being full and faithful.
