Does $\mathrm{Inn}(G) = \mathrm{Aut}(G)$ imply $Z(G) = \{e\}$? Can $|G| > |\mathrm{Aut}(G)|$? 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 $x\mapsto g\,x\,g^{-1}$.
 A: For your second question, take $G=\mathbb{Z}$. Then $Aut(G) \cong C_2$. For a finite case take $G = C_p$, a cyclic group of prime order $p$. Then $Aut(G) \cong C_{p-1}$.
A: 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 $2 \cdot 2^3 \cdot 2^3 \cdot 2^3 : 7$?
A: For the first question, you could take $G = S \times C_2$, where $S$ is a nonabelian simple group with trivial outer automorphism group. For example, $S=M_{11}$.
A: Nice to know (more than the OP asked for but strongly related): in group theory a group $G$ is called complete if both $Aut(G)=Inn(G)$ and $Z(G)=1$. An example is $G=S_3$. Theorem If a $N$ is a normal complete subgroup of a group $G$, then it is a direct factor of $G$.Proof (sketch) In general for $N \unlhd G$, there is a homomorphism $G/NC_G(N) \hookrightarrow Out(N):=Aut(N)/Inn(N)$, so $G=NC_G(N)$. And $N \cap C_G(N)=Z(N)=1$, whence $G=N \times C_G(N)$.
A: The kernel of this map is the center of $G$, so you demand that $G$ have a non-trivial center and no outer automorphisms. 
I think that $G = S_3 \times \mathbb{Z}/2$ does the job. Clearly it has a center (the second factor). Any automorphism fixes the center so induces an automorphism of $S_3$ which is inner.
EDIT: above is wrong (see comments). I am starting to believe the answer could be negative. For a $p$-group it is never true.
EDIT 2: well, now that I think about it, $G = \mathbb{Z}/2$ answers OP's original question, although not in a very satisfying way. 
