Q:

Outer automorphisms of the infinite symmetric group

Denote by S$_\infty$ the group of permutations of $\mathbb N$.
Question: Does there exist an outer automorphism of S$_\infty$, and if so, can one be exhibited? Does this depend on the continuum hypothesis?
Motivation: I'm trying to see whether the group of finitary permutations is characteristic in S$_\infty$.

A:

$S_{\infty}$ does not have any non-trivial outer automorphisms.
To prove this, we'll show that (1) the automorphisms of $S_{\infty}$ map transpositions to transpositions and (2) that this implies that $S_{\infty}$ has no non-trivial outer automorphisms.
To prove (1), let's consider the the conjugacy classes of $S_{\infty}$.  In the case of $S_n$, the fact that every permutation can be written as a product of disjoint cycles implies that the conjugacy classes correspond to the integer partitions of $n$.  In the case of $S_{\infty}$, we get a similar result.  Namely, the conjugacy classes of $S_{\infty}$ correspond to the multisets that contain elements in $\mathbb{N} \cup \{\lvert\mathbb{N}\rvert\}$.
In particular, we see that the transpositions form a conjugacy class.  Since every transposition has order $2$, we consider the other conjugacy classes of elements of order $2$.  The elements of order $2$ in $S_{\infty}$ are those that can be expressed as products of disjoint transpositions.  (We might need to take a product countably infinitely many disjoint transpositions, but this isn't a problem since it's easy to define infinite products of disjoint permutations.)  The conjugacy classes whose elements have order $2$ therefore correspond to pairs containing the number of transpositions in the product and the number of fixed points (which are both either positive integers or $\lvert\mathbb{N}\rvert$).  Let $\mathcal{C}$ denote the collection of all such conjugacy classes.
Now, the transpositions have the property that the product of two transpositions always has order at most $3$.  We claim that if $C \in \mathcal{C}$ is not the set of transpositions, then it contains a pair of elements whose product has order at least $6$.  Let $m \in \mathbb{N} \cup \{\lvert\mathbb{N}\rvert\}$ be the number of disjoint transpositions needed to express elements of $C$ as a product and let $n \in \mathbb{N} \cup \{\lvert\mathbb{N}\rvert\}$ be the number of fixed points.  The case where $m = 1$ corresponds to the transpositions.  Suppose $m \geq 2$.  If $n \geq 3$, then $C$ contains the elements
$$\pi_1 = (12) (56) \sigma$$
and
$$\pi_2 = (34) (67) \sigma$$
where $\sigma$ is a product of $m - 2$ disjoint transpositions with the property that $\sigma(k) = k$ for $1 \leq k \leq 7$ and $\sigma$ has $n - 3$ fixed points.  Then
$$\pi_1 \pi_2 = (12)(34)(567)$$
which has order $6$.  If $n < 3$, then $m = \infty$.  We can think of $S_{\infty}$ as acting on the set $\mathbb{Z} \uplus A$ where $\lvert A \rvert = n$.  Then $C$ contains the elements
$$\pi = \prod_{i \in 2 \mathbb{Z}} (i, i + 1)$$
and
$$\pi_2 = \prod_{j \in 2 \mathbb{Z}} (j - 1, j)$$
and we have $\pi_1 \pi_2(2 k) = 2k - 2$ and $\pi_1 \pi_2(2 k + 1) = 2k + 3$ for $k \in \mathbb{Z}$ so $\pi_1 \pi_2$ has infinite order.  This proves that $C$ contains a pair of elements whose product has order at least $6$ as we claimed.
It follows that any automorphism of $S_{\infty}$ maps the conjugacy class that consists of the transpositions to itself.  This completes the proof of (1).
To prove (2), we start by noting that two transpositions are disjoint or equal if and only if they commute.  From (1), we know that each transposition $(1 i)$ is mapped to some transposition $(j k)$.  Let $\phi$ be an automorphism of $S_{\infty}$.  Then $(1 2)$ and $(1 i)$ commute if and only if $\phi((1 2))$ and $\phi((1 i))$ commute.  Therefore, there exist $a_{\phi}$ and a bijection $\pi_{\phi}' : \mathbb{N} \setminus \{1\} \rightarrow \mathbb{N} \setminus \{a_{\phi}\}$ such that, for all $i \not= 1$, $\phi((1 i)) = (a_{\phi}, \pi_{\phi}'(i))$.  Equivalently, there exists $\pi_{\phi} \in S_{\infty}$ such that for all $i \not= 1$,
$$\phi((1 i)) = (\pi_{\phi}(1), \pi_{\phi}(i)) = \pi_{\phi} (1 i) \pi_{\phi}^{-1}$$
This implies that 
$$\phi(\sigma) = \pi_{\phi} \sigma \pi_{\phi}^{-1}$$
where $\sigma$ is finitary.
Next, we will show that $$\phi[\mathrm{Sym}(\mathbb{N} \setminus \{i\})] = \mathrm{Sym}(\mathbb{N} \setminus \{\pi_{\phi}(i)\})$$ for all $i$.  First, we observe that $C(S_3) = \mathrm{Sym}(\mathbb{N} \setminus \{1, 2, 3\})$ and by the above $\phi[S_3] = \mathrm{Sym}(\{\pi_{\phi}(1), \pi_{\phi}(2), \pi_{\phi}(3)\})$.  Since $C(\mathrm{Sym}(\{\pi_{\phi}(1), \pi_{\phi}(2), \pi_{\phi}(3)\})) = \mathrm{Sym}(\mathbb{N} \setminus \{\pi_{\phi}(1), \pi_{\phi}(2), \pi_{\phi}(3)\})$, we see that $\phi[\mathrm{Sym}(\mathbb{N} \setminus \{1, 2, 3\})] = \mathrm{Sym}(\mathbb{N} \setminus \{\pi_{\phi}(1), \pi_{\phi}(2), \pi_{\phi}(3)\})$.  Since $\langle (2 4), (3 4), \mathrm{Sym}(\mathbb{N} \setminus \{1, 2, 3\}) \rangle = \mathrm{Sym}(\mathbb{N} \setminus \{1\})$ and $\langle (\pi_{\phi}(2), \pi_{\phi}(4)), (\pi_{\phi}(3), \pi_{\phi}(4)), \mathrm{Sym}(\mathbb{N} \setminus \{\pi_{\phi}(1), \pi_{\phi}(2), \pi_{\phi}(3)\}) \rangle = \mathrm{Sym}(\mathbb{N} \setminus \{\pi_{\phi}(1)\})$, we see that
$$\phi[\mathrm{Sym}(\mathbb{N} \setminus \{1\})] = \mathrm{Sym}(\mathbb{N} \setminus \{\pi_{\phi}(1)\})$$
By taking conjugates by transpositions, we obtain
$$\phi[\mathrm{Sym}(\mathbb{N} \setminus \{i\})] = \mathrm{Sym}(\mathbb{N} \setminus \{\pi_{\phi}(i)\})$$
for all $i$ as claimed.
Now, let $\sigma \in S_{\infty}$.  We'll show that $\phi(\sigma) = \pi_{\phi} \sigma \pi_{\phi}^{-1}$.  Since
$$S_{\infty} / \mathrm{Sym}(\mathbb{N} \setminus \{i\}) = \left\{(i j) \mathrm{Sym}(\mathbb{N} \setminus \{i\}) \;\middle|\; j \in \mathbb{N}\right\}$$
it follows that for any $\tau \in S_{\infty}$, $\tau(i) = j$ if and only if $(i j) \tau \in \mathrm{Sym}(\mathbb{N} \setminus \{i\})$.
Consider $i \in \mathbb{N}$ and let $j = \sigma(i)$.  Then $(i j) \sigma \in \mathrm{Sym}(\mathbb{N} \setminus \{i\})$ so $\phi((i j) \sigma) \in \mathrm{Sym}(\mathbb{N} \setminus \{\pi_{\phi}(i)\})$.  Since $\phi((i j) \sigma) = (\pi_{\phi}(i), \pi_{\phi}(j)) \phi(\sigma)$, it follows that $\phi(\sigma)(\pi_{\phi}(i)) = \pi_{\phi}(j)$ so
$$\phi(\sigma)(\pi_{\phi}(i)) = \pi_{\phi} \sigma \pi_{\phi}^{-1}(\pi_{\phi}(i))$$
Since $i$ was arbitrary, it follows that $\phi(\sigma) = \pi_{\phi} \sigma \pi_{\phi}^{-1}$ as desired.

