Let G be a group. It looks to me as though you're forgetting the operation that makes Aut(G), the set of automorphisms of G, into a group -- namely composition of the automorphisms. So for x in G, define the automorphism I(x) of G to be conjugation by x. It is straightforward that the mapping I from G into Aut(G) is a homomorphism; just use the "product" in Aut(G), namely composition. So G/Ker(I) is isomorphic to the image, Inn(G). Here Ker(I) is easily seen to be the center Z(G) (the set of elements of G that commute with everybody). So always G/Z(G) is isomorphic to Inn(G) via the mapping I defined above. In particular, for S3, Z(S3) is the identity subgroup, so the isomorphism you're looking for is just I.