The symmetric group $S_n$ can be decomposed as a semi-direct product of the alternating group $A_n$ and a subgroup of $S_n$ of order 2. Use this fact to find a non-trivial outer automorphism of $A_n$ for $n \ge 3$.
Taking $S_n \cong A_n \rtimes \mathbb{Z}_2$ it would seem obvious that Out$(A_n)= \mathbb{Z}_2$. I'm not sure how to find the actual outer automorphism, however.
Using the above hypothesis, what is a non-trivial outer automorphism of $A_n$ for $n \ge 3$?
For $n\ge 3$ and $n\neq 6$ the non-trivial outer automorphism of $A_n$ is conjugation by an odd permutation, and we have $Out(A_n)\simeq C_2$. For $n=6$ we have $Out(A_6)\simeq C_2\times C_2$.
Edit: The question, why this outer automorphism of $A_n$ is not inner, has been answered here.