Looking on the Wikipedia page for automorphism; in the examples it first states that in set theory, the automorphism of a set $X$ is an arbitrary permutation of the elements of $X$, and these form the automorphism group, also known as the symmetric group, on $X$. However on the page Automorphisms of the symmetric and alternating groups, $ \operatorname{Aut}(S_n) = S_n $ except in the cases where $n=1,2,6$. So is the statement on the automorphism page incorrect when it says that all automorphism groups on $X$ are also known as symmetric groups, as not all of them are when for example $X=S_n$?
[Math] Automorphism groups and symmetric groups
abstract-algebragroup-theoryterminology
Related Solutions
First, on the permutation group page, there's this line "the term permutation group is usually restricted to mean a subgroup of the symmetric group."
Second, Cayley's theorem doesn't really make the terminology "permutation group" redundant. When you talk about a permutation group, I think you are implicitly giving an action of the group on a set of some objects. This is an extra data other than the group structure.
Lemma. Let $\mathrm{Aut}(G)$ be the automorphism group of a group $G$ and $\mathrm{cl}(G)$ the conjugacy classes of elements of $G$. Then $\mathrm{Aut}(G)$ acts on $\mathrm{cl}(G)$.
Explanation. Let $\alpha$ be an automorphism of $G$ and $\mathrm{cl}(a)=\{gag^{-1}:g\in G\}$ the conjugacy class of $a$. Then $\alpha(\mathrm{cl}(a))=\{\alpha(g)\alpha(a)\alpha(g)^{-1}:g\in G\}=\{t\alpha(a)t^{-1}:t\in G\}=\mathrm{cl}(\alpha(a))$ is also a class.
Lemma. If $\alpha\in\mathrm{Aut}(S_n)$ stabilizes the conjugacy class of transpositions then $\alpha\in\mathrm{Inn}(S_n)$ is inner.
Proof. We want to exhibit an element $\sigma\in S_n$ such that $\alpha(g)=\sigma g\sigma^{-1}$ for every $g$. It suffices to find a $\sigma$ for which $\alpha(\tau)=\sigma \tau\sigma^{-1}$ holds for every transposition $\tau$, since every permutation is a product of transpositions. It further suffices to find a $\sigma$ for which $\alpha(1k)=\sigma(1k)\sigma^{-1}$ for each $1< k\le n$, since these $n-1$ transpositions generate all the others.
Pick distinct $1<\ell_1,\ell_2\le n$. Then $\rho=(1\ell_2\ell_1)=(1\ell_1)(1\ell_2)$ has order $3$, in which case $\alpha(1\ell_1)\alpha(1\ell_2)$ has order three as well. Without loss of generality, $\alpha(1\ell_1)=(ab)$ and $\alpha(1\ell_2)=(ac)$ for some $a,b,c$; since this holds for each distinct pair $\ell_1,\ell_2$, we find that for every $k$, $\alpha(1k)=(af(k))$ for some $f(k)$.
Let $\sigma$ send $1$ to $a$ and $k$ to $f(k)$ for each $k>1$. This permutation satisfies our requirements.
Lemma. For $n\ne6$, the class of transpositions is the unique class of involutions with ${n\choose2}$ elements.
Proof. Consider the class of involutions with cycle type a product of $k>1$ disjoint cycles. Suppose
$$\frac{1}{k!}{n\choose 2}{n-2\choose2}\cdots{n-2(k-1)\choose2}={n\choose2}.$$
The LHS is the size of the new conjugacy class. Setting $m=n-2$ and $k=r+1$, cancelling ${n\choose2}$s,
$${m\choose2}\cdots{m-2(r-1)\choose2}=(r+1)!$$
The LHS is telescoping. Multiply by $2^r/(2r)!$ and get
$${m\choose2r}=\frac{(r+1)!2^r}{(2r)!}=\frac{r+1}{(2r-1)!!}$$
For $r>2$ the RHS is not even an integer (it is $<1$). The case $r=1$ gives $m(m-1)=4$ with no solution in $m$. Finally $r=2$ gives $m(m-1)(m-2)(m-3)=24$ which has positive solution $4$.
Theorem. For $n\ne6$, $\mathrm{Out}(S_n)=1$.
Proof. Combine the three lemmas. When $n\ne6$, automorphisms permute conjugacy classes and in particular stabilize the transposition class (since the size of this class is unique among the classes), hence every automorphism is inner.
Remark. This proof is adapted from this sketch from Wikipedia and these notes.
Best Answer
Automorphism of a set is an arbitrary permutation of its elements. An automorphism of a group is permutation of its elements which preserves the operation, i.e. $\varphi(xy)=\varphi(x)\varphi(y)$. Since every group $G$ is a set, you can look at two possible automorphism groups: one - $\operatorname{Aut}_{Set}(G)$ as a set and the other $\operatorname{Aut}_{Gp}(G)$ as a group. Cleraly $\operatorname{Aut}_{Gp}(G)\leq \operatorname{Aut}_{Set}(G)$, but usually they are not equal.
When talking about groups, the notion $\operatorname{Aut}(G)$ means the set of group-automorphism of $G$.
In your case, $\operatorname{Aut}(S_n)$ denotes the group-automorphisms of $S_n$, so there is no contradiction with the previous statements. BTW, there is no problem with $n=1$, since $S_1=\{id\}$ and $\operatorname{Aut}(S_1)=\{id\}$. To illustrate the problem in $S_2=\{id,(1,2)\}$, observe that there are two automorphisms of $S_2$ as a set: $$\begin{array}{c}id\mapsto id\\ (1,2)\mapsto(1,2)\end{array} \hspace{10pt} {\rm{and}}\hspace{10pt}\begin{align*}id &\mapsto (1,2)\\ (1,2)&\mapsto id\end{align*}$$ but the right automorphism is not a group-automorphism, hence there exists only one group-automorphism of $S_2$.