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What is the automorphism group of the symmetric group of degree three? From what I understand, the automorphism group of $Sym(n)$, with $n \neq 2$ or $6$ is 'trivial'. But I don't understand what trivial means in this case, is it that the automorphism group of $Sym(n) =$ the identity element?

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The outer automorphism group of $S_n$ for $n\neq 6$ is trivial. Any non-abelian group has non-trivial automorphisms: for any $g$ not in the centre, $G\rightarrow G$, $h\mapsto ghg^{-1}$ is a non-trivial automorphism. These automorphisms, i.e. those given by conjugation by some element of $G$, are called inner. The outer automorphism group is the group of all automorphisms modulo the normal subgroup of the inner ones.

Thus, for $n\neq 6$, all automorphisms of $S_n$ are those induced by conjugation by elements of $S_n$, as above.

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I want to add some information to @Alex's answer. It seems the word trivial you noted has another meaning because of:

Corollary: If $n\neq2$ or $n\neq6$ then $\text{Aut}(S_n)\cong S_n$.

So an automorphism, say $\phi$, of $S_n$ preserves transpositions iff it is inner and that's why $S_n$ is Complete. Moreover $\text{Aut}(S_3)\cong S_3\cong\text{Aut}(V)$, where in $V$ is the Klien group.

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@csss: Have you got your answer? – Babak S. Dec 9 '12 at 18:50

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