Does there exist a finitely generated group $G$ with outer automorphism group $\mathrm{Out}(G)$ finite, whose center contains infinitely many elements of order $p$ for some prime $p$?
A motivation is that the existence of such a group would answer the question in this 2011 MO post. Indeed such a group would be an example of a finitely generated group with finite Out but with finite index in a group with infinite Out (namely $G\times\mathbf{Z}/p\mathbf{Z}$).
Edit: here is an example when we drop the finite generation assumption: the universal central extension $G$ of $\mathrm{SL}_n(\mathbf{Q})$ when $n\ge 5$.
Indeed, the central kernel is isomorphic (by standard stability results for $K_2$ of fields) to $K_2(\mathbf{Q})$, which is isomorphic to $C_2\oplus\bigoplus_{p>2}C_{p-1}$, where
$p$ ranges over odd primes (see Milnor K-theory book, Chapter 11). So $K_2(\mathbf{Q})$ contains infinitely many elements of order 2 (or even of any other prime order, by Dirichlet's theorem of primes in an arithmetic progression). On the other hand, by Schreier-Van der Waerden, $\mathrm{Aut}(\mathrm{(P)SL}_n(K))$, for any field $K$, is generated by inner automorphisms, the inverse-by-transposition involution, and automorphisms induced by $\mathrm{Aut}_{\mathrm{field}}(K)$ acting entry-wise. Since here the field has a trivial field automorphism group, we obtain that $\mathrm{Out((P)SL}_n(\mathbf{Q}))$ has two elements only. It is immediate that the canonical map from $\mathrm{Aut}(G)$ to $\mathrm{Aut}(G/Z(G))$ is injective, and it follows that $\mathrm{Out}(G)$ is finite.
Actually I expect that central extensions of $\mathrm{SL}_n$ of some well-chosen finitely generated commutative ring should be a source of finitely generated exemples, but it sounds harder.
Best Answer
Theorem B of this paper implies that we can take any two nontrivial involution-free groups $A$ and $B$ and construct a complete simple groups $D$ with a (diagrammatically) aspherical presentation $D=A*B/\langle\!\langle w_1, w_2,\dots\rangle\!\rangle$ (though such use of this theorem is killing a mosquito with a cannon). Here, complete means naturally isomorphic to the automorphism group (i.e. centreless and without outer automorphisms). Now, suppose that $A$ and $B$ coincides with their commutator subgroups.
Acphericity implies that the centre of the free central extension $\widetilde D=A*B/[\langle\!\langle w_1, w_2,\dots\rangle\!\rangle,A*B]$ of $D$ is the free abelian group with the basis $\widetilde{w_1},\widetilde{w_2},\dots$ (see, e.g., Olshanskii's book, Section 34.4). Now, we can do whatever we want. For instance, we can take the quotient $G=\widetilde D/\langle\widetilde {w_1}^p,\widetilde{w_2}^p,\dots\rangle$ and obtain a desired group $G$. The natural map $\mbox{Aut}\,\widetilde D\to\mbox{Aut}\,G$ exists because the subgroup $\langle\widetilde {w_1}^p,\widetilde{w_2}^p,\dots\rangle$ is characteristic in $\widetilde D$ (as this subgroup consists of $p$th powers of all central elements), and this map is injective because $\widetilde D$ coincides with its commutator subgroup. So, there are no outer automorphisms of $G$.
Jul 6, 2015. More details added, as suggested by Mamuka.
(1) $G$ (as well as $D$ and $\widetilde D$) is finitely generated if $A$ and $B$ are finitely generated.
(2) Suppose that a group is a quotient of another group: $L=M/N$. Then
there is a natural map $f\colon \mbox{Aut}\,M\to \mbox{Aut}\,L$ if $N$ is characteristic;
$f(\mbox{Inn}\,M)=\mbox{Inn}\,L$ (i.e. $f$ sends inner automorphisms to inner automorphisms, and each inner automorphism of $L$ has at least one inner preimage);
$f$ is injective if $N$ is central and $M$ coincides with its commutator subgroup.
Now, take $M=G$ and $L=D$ (and $N=\langle\widetilde {w_1},\widetilde{w_2},\dots\rangle$). Then $\mbox{Aut}\,L=\mbox{Inn}\,L$ and, hence, $\mbox{Aut}\,M=\mbox{Inn}\,M$ by (2), i.e. all automorphisms of $M=G$ are inner.