Is $A_n$ characteristic in $S_n$? The title is the question. Is $A_n$ characteristic in $S_n$? 
If $\phi \in \operatorname{Aut}(S_n)$, Then $[S_n : \phi(A_n)]$ (The index of $\phi(A_n)$) is 2. 
Maybe the only subgroup of $S_n$ of index 2 is $A_n$?
Thanks in advance.

Let $H$ be a subgroup of $S_n$. Then either $H \subseteq A_n$ or $[H:H\cap A_n]=2$ (that is either $H$ is all even or half-even and half-odd). This fact can be proven by noticing that left multiplication by an odd permutation (if there is one in $H$) sends evens to odds and odds to evens bijectively (so there must be an equal number of both evens and odds if there are any odd elements to begin with).
Therefore, the only subgroup of index $2$ in $S_n$ is $A_n$ [If $H$ is all even and index $2$, it must be all of $A_n$. If $H$ is half-even and half-odd, then $A_n$ has a normal subgroup: $H \cap A_n$ of index 2 in $A_n$, but no such subgroup exists by inspection for $n=1,2,3,4$ and simplicity of $A_n$ for $n \geq 5$]. Thus $A_n$ is characteristic (being the unique subgroup of $S_n$ of order $n!/2$ it must be sent to itself by any automorphism).
