$\Aut(S_3)=\Inn(S_3)\cong S_3$

The center of \(S_3\) is trivial, so conjugation gives \[ \operatorname{Inn}(S_3)\cong S_3/Z(S_3)\cong S_3, \] and hence \(|\operatorname{Inn}(S_3)|=6\).

Any automorphism of \(S_3\) preserves element orders, so it permutes the three transpositions. Since two distinct transpositions generate \(S_3\), an automorphism is determined by the images of two fixed distinct transpositions. There are at most \(3\cdot2=6\) such choices. Therefore \[ 6=|\operatorname{Inn}(S_3)|\le |\operatorname{Aut}(S_3)|\le6. \] Thus \[ \operatorname{Aut}(S_3)=\operatorname{Inn}(S_3)\cong S_3. \]