Every homomorphism $A_n\to S_n$ extends to an endomorphism of $S_n$ for $n\geq 5$

Question

Let $n\geq 5$, $S_n$ the symmetric group on $n$ letters and $A_n$ the corresponding alternating group.

I want to show that every homomorphism $g:A_n\to S_n$ extends to an endomorphism $\tilde{g}:S_n\to S_n$ compatible with the inclusion $i:A_n\to A_n$, i.e. $\tilde{g}\circ i=g$.

Since, for $n\geq 5$ the group $A_n$ is simple, $g$ must be injective or trivial, so let us focus on the injective case. Since we need $\tilde{g}\circ i=g$, it follows that $\tilde{g}$ must be injective too. From groupprops I know that for $n\geq 5$ the elements of $End(S_n)$ are one of these three types: automorphisms, trivial, have image of order two.

Therefore, $\tilde{g}$ must be an automorphism. From the same page I know that for $n\neq 6$ we have $Aut(A_n)=Aut(S_n)=S_n$, all of them given by conjugation. Now, since $g$ is an isomorphism onto its image, my first question raises:

1. Are there subgroups of $S_n$ isomorphic to $A_n$ which are not equal to $A_n$ (defined as the subgroup of even permutations)? If not, then $g$ is an automorphism of $A_n$, which is given by conjugation by an element of $S_n$ and therefore can be easily extended to all $S_n$.

For the case $n=6$, I haven't been able to find the automorphism structure of $S_n$ and $A_n$, I only know that $S_n< Aut(S_n)=Aut(A_n)$. So my second question is:

2. How can I extend $g$ when $n=6$?

Answer 1

Suppose $H=g(A_n)\ne A_n$ so $H\cap A_n\ne A_n$.

Since $[S_n:H]=2$, $H\trianglelefteq S_n$ and therefore $H\cap A_n\trianglelefteq A_n$ contradicting the fact $A_n$ is simple.

Hence $g$ is an automorphism of $A_n$ and you know the rest.