Actually I have solved the part "$\operatorname{Inn}(S_3)\simeq S_3$" by using the result-
"Let $G$ be a group. Then ${G \over Z(G)}\simeq \operatorname{Inn}(G)$, where $Z(G)=\{g\in G\mid gx=xg\: \forall x \in G\}$"
So I put $G=S_3$ and hence $Z(S_3)=\{e\}$, where $e$ is identity permutation.
And thus ${S_3 \over Z(S_3)}\simeq S_3$
Finally, using the above result, we get, $\operatorname{Inn}(S_3)\simeq S_3$.
But how to prove $\operatorname{Aut}(S_3) \simeq S_3$. Although, One thing I have notice that, to prove $\operatorname{Aut}(S_3) \simeq S_3$, it is enough to show that $\operatorname{Inn}(S_3)\simeq \operatorname{Aut}(S_3)$. But I am stuck with it also.
Can anyone suggest me a clear and rigorous wayout?
Thanks for your help in advance.
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