more on outer automorphisms of $S_6$ A sequel to my earlier question:
I've been doing some more concrete arithmetic with one of these outer automorphisms and it's working out just the way "they" say it will (e.g. $(123)(45)(6)$ goes to $(ABCDEF)$ and $(123456)$ goes to $(A)(BC)(DEF)$, etc. etc.).
My next question: If I'm not mistaken, $S_6$ should be a subgroup of some larger group of permutations, but smaller than the group of all permutations on that larger set---I would think a group of permutations on a set that has $\{1,2,3,4,5,6\}$ as a subset---such that some inner automorphism of that larger group, when restricted to $S_6$, is an outer automorphism of $S_6$.  How big is that larger set and what group of permutations on it do we need to look at, and which of its members do we conjugate by?
 A: As noted by Bill Cook, there is a natural group to look at, which is the holomorph of $S_6$. 
This works for any group.
More generally: if $N$ is a group, and $\varphi\colon H\to \mathrm{Aut}(N)$ is a group homomorphism, then this induces an action of $H$ on $N$ by the formula 
$${}^hn = [\varphi(h)](n),$$
where "${}^hn$" means "the image of $n$ under the action of $h$". The formula makes sense because $\varphi(h)\in\mathrm{Aut}(N)$, so we can evaluate $\varphi(h)$ on $n$.
This action satisfies nice properties:


*

*${}^h(nm) = {}^hn{}^hm$ for all $n,m\in N$, $h\in H$.

*${}^k({}^hn) = {}^{kh}n$. 


Given such a triple, $(N,H,\varphi)$, we can construct a $G$ that contains subgroups $\mathcal{N}$ and $\mathcal{H}$ isomorphic to $N$ and $H$, respectively, with $\mathcal{N}\triangleleft G$, $G=\langle\mathcal{N},\mathcal{H}\rangle=\mathcal{NH}$, and where for every $\mathfrak{h}\in \mathcal{H}$ and $\mathfrak{n}\in\mathcal{N}$, if $h\in H$ corresponds to $\mathfrak{h}$ and $n$ corresponds to $\mathfrak{n}$, then $\mathfrak{hnh}^{-1}$ corresponds to ${}^hn$. This is the semidirect product $N\rtimes_{\varphi}H$. 
The underlying set of $N\rtimes_{\varphi}H$ is $N\times H$. The multiplication rule is 
$$(n_1,h_1)\cdot (n_2,h_2) = \Bigl(n_1{}^hn_2,h_1h_2\Bigr).$$
I'll leave it to you to verify this group satisfies the properties given above, with 


*

*$\mathcal{N} = \{ (n,1)\mid n\in N\}$;

*$\mathcal{H} = \{(1,h)\mid h\in H\}$. 


Note that, indeed, we have:
$$(1,h)(n,1)(1,h)^{-1} = (1{}^hn,h)(1,h^{-1}) = ({}^hn,h)(1,h^{-1}) = ({}^hn{}^h1,hh^{-1}) = ({}^hn,1).$$
Conversely, if $G$ is a group, $N\triangleleft G$, $H\leq G$ are subgroups with $NH=\langle N,H\rangle = G$ and $N\cap H=\{1\}$, then for each $h\in H$ we have an automorphism of $N$ given by $n\mapsto hnh^{-1}={}^hn$; this yields a group homomorphism $\varphi\colon H\to \mathrm{Aut}(N)$, and it is easy to verify that the group $N\rtimes_{\varphi}H$ constructed as above is isomorphic to $G$. 
Now, by taking $\varphi\colon\mathrm{Aut}(G)\to\mathrm{Aut}(G)$ being the identity, we can always construct the group $G\rtimes_{\mathrm{id}} \mathrm{Aut}(G)$, which is the aforementioned holomorph of $G$. 
A: As others have pointed out, you can embed any group in its holomorph. But for a group $G$ with trivial centre (like $S_6$), the normal subgroup ${\rm Inn}(G)$ of ${\rm Aut}(G)$ is naturally isomorphic to $G$, so you can embed $G$ directly into ${\rm Aut}(G)$, which is of course a smaller group than the holomorph.
In fact ${\rm Aut}(S_6)$ embeds into $S_{12}$ as follows. Let $\tau$ be an outer automorphism of ${\rm Aut}(S_6)$ with $\tau^2=1$. For an element $g \in S_6$ acting on the set $\{1,2,3,4,5,6\}$, let $\tau'(g)$ denote $\tau(g)$, but acting on the set $\{7,8,9,10,11,12\}$. So, for example, if $g=(1,2)$ and $\tau(g) = (1,2)(3,4)(5,6)$, then $\tau'(g) = (7,8)(9,10)(11,12)$.
Then we can embed $S_6$ into $S_{12}$ by $g \to g\ \tau'(g)$. Then $t := (1,7)(2,8)(3,9)(4,10)(5,11)(6,12)$ induces $\tau$ by conjugation on the image of this embedding, so $\tau$ together with this image generates ${\rm Aut}(S_6) \le S_{12}$.
This image is a subgroup of the Mathieu group $M_{12}$ and can be used as part of one of the many constructions of $M_{12}$.
