How does one go about proving that a particular group is isomorphic to the set of its inner automorphisms? Specifically I'm looking at the symmetric group of order 3. I've done the tedious calculations conjugating the entire group by each element, and indeed there are 6 unique mappings from S_{3 }to Inn(S_{3}), but so what?

More specifically, here's what I've done:

Denote:

1 as the identity which fixes each element (1)(2)(3)

2 is the cycle (123)

3 is the cycle (132)

4 is (12)

5 is (13)

6 is (23)

The inner automorphism induced by the identity gives us the permutation (1)(2)(3)(4)(5)(6)

From this permutation, the other automorphisms are:

Induced by 2: (645)

3: (564)

4: (23)(465)

5: (23)(654)

6: (23)(546)

Aside from routine calculation showing that conjugation is one-to-one, onto, and operation preserving, I have no other work on this.

I feel like I can see the isomorphism laying here, I just don't know what bits and pieces I need to assemble in a proof. Or maybe I'm not even close?

Thanks

Anthony