Finding a group such that the map $\varphi: G \to \mathrm{Aut}(G)$, $g \mapsto f_g$ (conjugation by $g$) is an isomorphism I know that the map $\varphi: G \to \mathrm{Aut}(G)$ sending an element $g$ to the map conjugation by $g$, which I will denote by $f_g$, is a homomorphism. I'm trying to find a $G$ such that the map is also an isomorphism.
I know from my reading on the subject that $S_3$ works, but I'm trying to fully understand how I would come up with this on my own with being prompted to consider $S_3$. The first important thing is this map $\varphi$ would have to be injective (in order to be an isomorphis), so that would imply it would have to have trivial kernel: for $g \in G$,
\begin{align*}
a \in \mathrm{ker}(\varphi) & \iff \varphi(a) = \mathrm{id}_G \\
& \iff \forall a \in G, \; c_a = \mathrm{id}_G \\
& \iff \forall a \in G, \forall x \in G, \; c_a (x) = x \\
& \iff \forall a \in G, \; axa^{-1} = x \\
& \iff \forall x \in G, \; ax = xa \\
& \iff a \in Z(G).
\end{align*}
So if $\phi$ is an isomorphism, it is injective, so its kernel is trivial. But we just showed that $\mathrm{ker}(\varphi) = Z(G)$, so $Z(G)$ is also trivial. So it seems that this is one requirement for $G$ to be an isomorphism.
For $\phi$ to be an isomorphism, it must also be the case that $|G| = |\mathrm{Aut}(G)|$. I think this implies that the entire automorphism group of $G$ must be inner automorphisms. Is that correct? For every $g \in G$, $c_g$ is an automorphism and an "inner" automorphism. At least in the case of finite groups, I only have $|G| = |\mathrm{Aut}(G)|$ if there are no outer automorphisms. Is is true in general that $\mathrm{Aut}(G) =\mathrm{Inn}(G)$ if and only if the center of $G$ is trivial?
If not, how else would I come up with a group like $S_3$?
 A: It is not true in general that $Z(G) = 1$ implies ${\rm Aut}(G) = {\rm Inn}(G)$.
There are many examples, for example $G = A_4$ has $Z(G) = 1$, but $[{\rm Aut}(G) :{\rm  Inn}(G)] = 2$. In fact ${\rm Aut}(G) \cong S_4$ in this case. (For an example of an outer automorphism of $G$, take $f: G \rightarrow G$ defined by $f(x) = gxg^{-1}$ where $g = (12)$ is a transposition from $S_4$.)
In general describing automorphism groups is not always so easy.
For the question of how you would come up with $G = S_3$ as an example, as you note you should look for groups with $Z(G) = 1$. A good start for any problem is to look at some small examples, and it turns out in this case the smallest example works.
A: Yes, because the kernel of the homomorphism from $G$ to $\rm {Aut}(G)$ given by $\varphi (g)=i_g$ (the inner automorphism $x\mapsto gxg^{-1}$) is the center $Z(G)$.  So by the first isomorphism theorem,  $G/Z(G)\cong \rm{Inn}(G)$. So when the center is trivial,  $G\cong\rm{Inn}(G)\le\rm{Aut}(G)$.
For $n\gt2$, $S_n$ is centerless (fairly easy).

Also,  other than $n=2,6$, all the automorphisms of $S_n$ are inner.   Putting it together,  we get that $S_n\cong\rm{Inn}(S_n)\cong\rm{Aut}(S_n)$.  Such groups are called complete.
