TITLE: Does $\mathrm{Inn}(G) = \mathrm{Aut}(G)$ imply $Z(G) = \{e\}$? Can $|G| > |\mathrm{Aut}(G)|$?
QUESTION [6 upvotes]: Is there a group $G$ whose conjugation map1 $G \to \mathrm{Aut}(G)$ is surjective without being injective?
More generally, is there a group $G$ (not necessarily finite) such that $|G| > |\mathrm{Aut}(G)|$?

1 I.e. the map sending every $g\in G$ to the automorphism of $G$ defined by $x\mapsto g\,x\,g^{-1}$.

REPLY [4 votes]: For the first question, here are the smallest three examples:

$G=C_2$
$G=C_2 \times \operatorname{A\Gamma L}(1,8)$
$G\cong 2\cdot 2^3 \cdot 2^3 : 7$ is the normalizer of a Sylow 2-subgroup in a double cover of Suzuki's simple group of order 29120

The second one generalizes to $C_2 \times \operatorname{A\Gamma L}(1,2^{2n+1})$. The latter direct factor is a complete group with no normal subgroups of index 2. Derek Holt mentions $C_2 \times M_{11}$ as another such group.
The first and third ones are neat because they are directly indecomposable.
I am interested in more indecomposable examples.
I know there are no such examples that are $p$-groups other than $C_2$. I find it odd that $2\cdot 2^3 \cdot 2^3 : 7$ exists, but $2 \cdot 2^3 : 7$ does not. Is there a $2 \cdot 2^3 \cdot 2^3 \cdot 2^3 : 7$?