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Let $G$ be a group, $X$ a generating set of $G$. Suppose $X$ is $\operatorname{Aut}(G)$-invariant, i.e. $\sigma(X)\subseteq X$ for all $\sigma \in \operatorname{Aut}(G)$. When is the restriction homomorphism $$ \begin{gathered} \operatorname{Aut}(G) \to \operatorname{Sym}(X)\\ \sigma \mapsto \sigma|_X \end{gathered}$$ an isomorphism?

Example: $A_4$ with $X=\{\text{3-cycles}\}$ satisfies the above properties. $C_2\times C_2$ works, and so do all cyclic groups of order up to 4 (Thanks Jeremy Rickard for the correction). Are there any other examples?

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  • $\begingroup$ I don't think $A_4$ is an example. Cyclic groups of order up to $4$ and $C_2\times C_2$ are. $\endgroup$ – Jeremy Rickard Oct 2 '14 at 20:58
  • $\begingroup$ @JeremyRickard $\operatorname{Aut}(A_4)=S_4$, and each automorphism takes takes 3-cycles to 3-cycles. See groupprops.subwiki.org/wiki/… $\endgroup$ – Avi Steiner Oct 2 '14 at 21:01
  • $\begingroup$ But there are eight $3$-cycles. $\endgroup$ – Jeremy Rickard Oct 2 '14 at 21:03
  • $\begingroup$ @JeremyRickard ... oops! You're right. $\endgroup$ – Avi Steiner Oct 2 '14 at 21:04
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    $\begingroup$ Going back to the question, I think that for $n\ge 5$ there's nothing not abelian, because then every orbit of $Alt(n)$ on $Sym(n)-\{1\}$ has cardinal $>n$. Hence I guess that the answer for your question is a short finite list. $\endgroup$ – YCor Oct 2 '14 at 21:56
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For $G$ abelian, the complete list is $C_2$, $C_3$, $C_4$, $C_6$ and $C_2\times C_2$.

Proof: Suppose that $X$ contains a non-involution $x$. Then $x^{-1}\in X$ (since inversion in an automorphism of $G$) but then $\{x,x^{-1}\}$ is a block of size 2 for the action of Aut(G) on $X$ and thus $|X|=2$ and $X=\{x,x^{-1}\}$ and $G$ is cyclic. Since $X$ is $Aut(G)$-invariant, $G$ must have only two element of order $|G|$ and this happens only when $G$ is $C_3$, $C_4$ or $C_6$. We may thus assume that $G$ is generated by involutions and hence is an elementary abelian $2$-group. It's easy to check that $C_2^n$ has this property only for $n=1$ and $n=2$.

(I was assuming $G$ is finite, but the proof can probably be adapted.)

For $G$ nonabelian, I think the only example is $G=Sym(3)$ with $X$ the set of involutions. You could probably prove this following YCor's comments for example.

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