Automorphism group of alternating group.

Question

As we know $\operatorname{Aut}(A_n)\cong \operatorname{Aut}(S_n)\cong S_n$ for $n\geq4$, $n\neq 6$ and $\operatorname{Aut}(A_6)\cong \operatorname{Aut}(S_6)\cong S_6\rtimes C_2$.

1. Is there any method to obtain an element $g$ in $S_n$ ($n\neq 6$) for $f\in \operatorname{Aut}(S_n)$ so that $f$ is the conjugation map by $g$?

2. Another problem is that if $H$ is subgroup of $A_6$ and $\sigma\in\operatorname{Aut}(A_6)$ then are the subgroups $H$ and $\sigma(H)$ conjugate in $S_6$? Also what can we say about $H$ and $\sigma(H)$ if $n\neq 6$?

Answer 1

For the first question, compute $f$ on the $n$-cycle $(12\dots n)$ to obtain $(a_{1} a_{2} \dots a_{n})$. Then compute $f$ on the $2$-cycle $(12)$ to obtain $(b_{1} b_{2})$. Now within the $a_{i}$ there are two consecutive ones (indices are taken modulo $n$) such that

1. $a_{i} = b_{1}, a_{i+1} = b_{2}$, or
2. $a_{i} = b_{2}, a_{i+1} = b_{1}$.

Your $g$ is $$ 1 \mapsto a_{i}, 2 \mapsto a_{i+1}, \dots , n \mapsto a_{n+i-1}, $$ where here, too, indices are taken modulo $n$.

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Explanation Recall that $$ (12\dots n) = (1^{g} 2^{g} \dots n^{g}), $$ where $i^{g}$ indicates the image of $i$ under the permutation $g$. So if $f$ is the same thing as conjugation by $g$, computing the image of $(12\dots n)$, under $f$ will give you one of $$ (1^{g} 2^{g} \dots n^{g}) = (2^{g} \dots n^{g} 1^{g}) = (3^{g} \dots n^{g} 1^{g} 2^{g}) = \dots $$ To see what the images of $1$ under $g$ is, compute $f$ on $(12)$.

Answer 2

For second question, I think it is not true. for example take $H=\langle (1,2,3)\rangle$ and there is a $\sigma \in \operatorname{Aut}(A_6)$ such that $\sigma(H)=\langle (1,2,3)(4,5,6)\rangle$ but $H$ and $\sigma(H)$ are not conjugate. And for third question, I think it is true because all automorphisms of $A_n$ ($n\neq 6$) are conjugation by elements of $S_n$. (Please correct if there is any mistake).