Number of Automorphisms of $S_6$ Trivial question about counting the number of automorphisms of $S_6$:
I know that for all $n \geq 3$, $Z(S_n)=1$, so Inn$(S_n) \cong S_n$.
I also know that $S_6$ has nontrivial outer homomorphisms, Out$S_n \cong \mathbb{Z}_2$.
Does this mean there are $S_n + \mathbb{Z}_2$ automorphisms in total, since Inn$(S_n)$ and Out$(S_n)$ are disjoint sets? So, other symmetric groups (excluding $S_2$) have $n!$ elements, while $S_6$ has $6! + 2$.

$\mathrm{Out} (S_n) $ is not a group of automorphisms, but is the quotient of the automorphism group by the subgroup of inner automorphisms. Thus 
$$|\mathrm{Out}(S_n) |=|\mathrm{Aut} (S_n) |/|S_n|$$
so you can find the order of the automorphism group this way.
Essentially an outer automorphism is an equivalence class of automorphisms $f(x) $ where $f(x) \sim \sigma f(x) \sigma^{-1}$ for any $\sigma$. 
