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Composing two bijections is a bijection (when domains and codomains match up), and composing two homomorphisms is a homomorphism, so $\text{Aut}(G)$ is closed under associative function composition. And $g \mapsto g$ is an automorphism, and the inverse of an isomorphism is also an isomorphism, so $\text{Aut}(G)$ is a group. It is a subset of all the permutations of $G$ as a set, so if $|G| = n < \infty$, then $\text{Aut}(G) \leq S_n$. "The automorphism group is a subgroup of $S_n$."

However, we can't go the other way around. Given $S_n$, it doesn't make sense to say "this subgroup of $S_n$ is the automorphism group", because we need to specify "is an automorphism group with respect to..." some group $G$. Although, since there is only one automorphism group for a given group $G$, we could also say "is the automorphism group of $G$".

This distinction confused me a lot when I first thought about it. It also leads to a question: is there a "nice" description of each subgroup of $S_n$ (say, each subgroup of $S_6$) as the automorphism group of some well-known group? Of course, we will need a few different groups to accomplish this.

Related: Realizable automorphism groups of graphs

EDIT: Based on the comments, this question is very general and difficult. To make it answerable, I will narrow the scope. Are there any algorithms, theorems, or insights related to the following problem? Solving it is too much to ask, but I'd be happy to hear how one might begin to approach this task.

Given a finite group $H$ of reasonably small order, find a group $G$ such that $H \cong \text{Aut}(G)$.

jskattt797
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    https://math.stackexchange.com/questions/253936/is-every-group-the-automorphism-group-of-a-group – Elle Najt Sep 13 '20 at 20:50
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    If $n\notin {2,3,6}$ then $\text{Aut}(A_n)=\text{Aut}(S_n)=S_n$. See, e.g., this – lulu Sep 13 '20 at 20:55
  • Good points! So we'd have to forget about describing any cyclic subgroups of odd order. And we can describe almost any symmetric group subgroup as the automorphism group of itself. – jskattt797 Sep 13 '20 at 21:01
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    Every finite group is a subgroup of some $S_n$, so are you just asking how to describe finite groups as automorphism groups of other finite groups? Or are you specifically asking whether a given subgroup of $S_n$ can be described as the automorphism group of a group of order $n$? – Qiaochu Yuan Sep 13 '20 at 21:22
  • Oh right. So asking the question for arbitrary $n$ might be too hard. But yes, I am wondering if, given a permutation group, there is a way to describe it as the automorphism group of some group. Sometimes this is not possible, but how would we go about it when it is possible? – jskattt797 Sep 14 '20 at 05:52
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    It is very much unclear what you are asking. You already know that some finite groups are not isomorphic to $Aut(G)$ for any $G$. A classification of finite groups which are isomorphic to $Aut(G)$ for some $G$ is a hard open problem. Are you asking for an algorithm? – Moishe Kohan Sep 14 '20 at 20:39
  • You could ask the question about outer automorphism groups: is every finite group the outer automorphism group of some group. This question stands a chance of being answerable :-) – user1729 Sep 15 '20 at 06:38
  • @user1729 This, is a theorem (due to Matumoto). – Moishe Kohan Sep 15 '20 at 16:58
  • @Moishe I had meant to write: "is every finite group the outer automorphism group of some finite group". I had read the question assuming that all the groups are finite, although I see now that this isn't necessarily the case. With this altered question, we cannot apply Matumoto's theorem, as I believe his groups are always infinite (given a finite group $H$ he gives you an infinite group $G$ with $\operatorname{Out}(G)\cong H$ - his groups $G$ are fundamental groups of graphs of groups, so always infinite). This question probably also has a known answer, but seems interesting anyway :-) – user1729 Sep 15 '20 at 18:33
  • @user1729 Oh, I see. I think, you should ask this as a separate question. – Moishe Kohan Sep 15 '20 at 20:42
  • @MoisheKohan Have done so now. – user1729 Sep 16 '20 at 09:56

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