The automorphism group of $S_6$ is isomorphic to a semidirect prodct On this document an outer automorphism of $S_6$ is constructed. I would like to use this construction to prove that $\mathrm{Aut}(S_6)\cong S_6\rtimes_\varphi\mathbb{Z}_2$. The idea would be to find an outer automorphism of order 2. If $F$ is the outer automorphism constructed in the document above then I can prove that the order of $F$ is even. Any outer automorphism is of the form $F\gamma_\sigma$ for some inner automorphism $\gamma_\sigma$. I can also prove that $F^2$ is inner so $F^2=\gamma_\tau$. I think one needs to use $\tau$ to construct $\sigma$ so that $F\gamma_\sigma$ has order 2. This is basically the approach used in this paper. But it seems to me the the construction of the outer automorphism done in the paper is a bit different than the construction I am interested in. I feel like there should be a way to modify Rotman's argument to make it work for the first construction. Any ideas?
 A: I don't see how Rotman's construction can easily be used to find an outer automorphism of order $2$. That method uses the action on the cosets of a transitive subgroup of $S_6$ that is isomorphic to $S_5$. But the resulting outer automorphism depends on the numbering of the six cosets, and there are $720$ possible numberings. So you are effectively just looking for the innter automorphism $\gamma_\sigma$ that you describe.
I think the easiest way to do this is just to write down an outer automorphism of order $2$. To define the automorphism it suffices to specify the images of the transpositions $(1,2),(2,3),(3,4),(4,5),(5,6)$, each of which should be the product of three disjoint transpositions.
From the Coxeter presentation of $S_6$, it follows that such an assignment will define an automorphism provided that the images of two generators $x,y$ commute if and only if $x$ and $y$ do. This makes it possible to find the sought after automorphism of order 2 without too much difficulty.
I did this and found the automorphism of of order 2 below. This was not so hard. I knew that the iamges of $(1,2)$, $(3,4)$ and $(5,6)$ must all commute, so I wrote those down first, then I just found the images of $(2,3)$ and $(4,5)$ by trial and error.
$(1,2) \mapsto (1,2)(3,4)(5,6)$;
$(2,3) \mapsto (1,6)(2,3)(4,5)$;
$(3,4) \mapsto (1,2)(3,5)(4,6)$;
$(4,5) \mapsto (1,4)(2,3)(5,6)$;
$(5,6) \mapsto (1,2)(3,6)(4,5)$.
