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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 $X=S_n$?

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Automorphism of a set is an arbitrary permutation of it's elements. An automorphism of a group is permutation of it's 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 $S_2$.

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Very clear explanation, thanks –  user50229 Nov 23 '12 at 15:18

In general, an automorphism of an object $A$ is a morphism $f\colon A\to A$ such that there exists a morphism $g\colon A\to A$ such that the composositions $f\circ g$ and $g\circ f$ are both the identity morphism of $A$.

To step from this abstract concept to some concrete situation, one must specify what kind of objects one talks about and what a morphism is. For sets (or as one says: in the category Set), an object is, of course, just a set and a morphism is simply a map. Then an automorphism of a set $A$ is simply an arbitrary bijective map $A\to A$.

On the other hand, for groups (in the category Grp), the objects are groups and the morphisms are group homomorphisms, i.e. maps that "respect the group law". For example, an automorphism of a group necessarily maps the identity element to itself, whereas a bijection (or "set-automorphism") need not obey this restriction.

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That's very helpful to know the general use of the term, thanks –  user50229 Nov 23 '12 at 15:17
    
I would also accept your answer, but unfortunately it only lets you select one? –  user50229 Nov 23 '12 at 15:19
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Rereading the answer, should it not be the compositions $f \circ B$ and $B \circ f$ that are both the identity morphism of $A$? –  user50229 Nov 23 '12 at 16:40
    
@SJGreen Thanks, I corrected. No idea why I wanted to call the inverse $B$ in the first place. :) –  Hagen von Eitzen Nov 23 '12 at 16:42

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