How general are $\mathrm{Aut}$ groups and $\mathrm{End}$ rings?

Question

For any locally small category $X$ and object $A\in X$ the set $\mathrm{Aut}_X(A)$ is a group w.r.t. composition $\circ$. For any locally small abelian category $X$ and object $A\in X$ the set $\mathrm{End}_X(A)$ is a ring w.r.t. $+,\circ$.

1. Does for any group $G$ exist $X$ and $A$ with $G\cong\mathrm{Aut}_X(A)$? We know that $G$ is a quotient of the free group $F_G$, and is a subgroup of the permutation group $S_G=\mathrm{Aut}_{Set}(G)$.

2. Does for any unital ring $R$ exist $X$ and $A$ with $G\cong\mathrm{End}_X(A)$? We know that $R$ is a quotient of the free ring $\mathbb{Z}\langle R|\emptyset\rangle$. Is it also a subring of some 'typical' ring? Perhaps $\mathrm{End}_{\mathbb{Z}}(?)$.

This post is related.

Answer 1

[Edit: my initial remarks were based on an answer I discarded: sorry for the confusion]

1. Let $G$ be an arbitrary group, and let $\mathsf{Set}(G)$ be the category of sets with (right) $G$-action. The group $G$ acts on itself by multiplication on the right, and it is easy to check that $\operatorname{End}_{\mathsf{Set}(G)}(G)=\operatorname{Aut}_{\mathsf{Set}(G)}(G)=G$, with $g\in G$ corresponding to $\lambda_g:G\to G$ given by $\lambda_g(h) = g h$.

2. This is essentially identical. Let $\mathsf{Mod}(R)$ be the category of right $R$-modules. It is easy to check that $\operatorname{End}_{\mathsf{Mod}(R)}(R)=R$, where $r\in R$ corresponds to $\lambda_r:R\to R$ given by $\lambda_r(s)=r s$.

Edit: The groups $\operatorname{Aut}_{\mathsf{Set}}(A)$ are all permutation groups (finite or infinite). On the other hand, I do not think groups of the form $\operatorname{Aut}_{\mathsf{Grp}}(A)$ or $\operatorname{Aut}_{\mathsf{Mod}(R)}(M)$ have been classified, except perhaps for very special cases of $R$ (e.g. if $R$ is a field then $\operatorname{Aut}_R(M)$ will be a general linear group over $R$).