Analogues of the dihedral group A virtually-$\mathbb{Z}$ group $G$ admits either a epimorphism onto $\mathbb{Z}$ or a epimorphism onto $D_\infty$. 
So what happens if one replaces $\mathbb{Z}$ by another group $F$ (like the free group or $\mathbb{Z}^n$). My first guess would be that any virtually-$F$ group $G$ maps surjectively onto one of the groups
$F\rtimes H$, where $H\le $Aut$(F)$ is any finite subgroup. 
This is clear in the case where $G$ is semidirect product of $F$ and a finite group $K$; one can simply take $H$ to be the image of $K\rightarrow $ Aut $ (F) $.
So my questions are:
1) Is it still true that there is an epimorphism even in the non-split case?
2) Is it also true that two groups $F\rtimes H_1$ and $F\rtimes H_2$ (with $H_i\subset $ Aut $(F)$ finite) cannot surject onto each other unless $H_1$ and $H_2$ are conjugated (in which case the groups are isomorphic) ?
I doubt that this is true for all groups $F$ but maybe one can find sufficient conditions that guarantee this.
 A: Here's an idea for a proof that the modular group $\Gamma=\mathbb{Z}/2*\mathbb{Z}/3$, which is, of course, virtually free, doesn't surject a group of the form $F\rtimes H$.  I don't have time to work out the details.
First, I think it's plausible that the only reduced, non-trivial graph-of-groups decomposition for $\Gamma$ is the one that realises it as the free product given above, which I will denote by $\mathcal{H}$.  One idea for the proof is as follows.  If $\mathcal{G}$ is any such graph of groups with fundamental group $\Gamma$, then the generators $a$ of order two and $b$ of order 3 can be taken to lie in distinct vertices of $\mathcal{G}$; this leads to an obvious immersion of graphs of groups $\mathcal{H}\to\mathcal{G}$.  But $\mathcal{H}$ carries the whole of $\Gamma$, so the immersion is actually an isomorphism.
Remark:  If the above is true, then it should be known, and there should be a reference out there somewhere.
Now suppose that $f:\Gamma\to F\rtimes H$ is a surjection.  Then $F\rtimes H$ can be decomposed as a graph of groups $\mathcal{J}$, and this induces a graph of groups decomposition $\mathcal{J}'$ for $\Gamma$, by setting $\mathcal{J}'_v=f^{-1}\mathcal{J}_v$ for each vertex $v$ and $\mathcal{J}'_e=f^{-1}\mathcal{J}_e$ for each edge $e$.  But, by the previous claim, $\mathcal{J}'=\mathcal{H}$.  As $\mathbb{Z}/2$ and $\mathbb{Z}/3$ are both simple, it follows that $f$ is an isomorphism. But $\Gamma$ is not of the form $F\rtimes H$.  (To see this, note, for instance, that all the torsion in $F\rtimes H$ is conjugate into a singe finite subgroup $H$; this is not the case in $\Gamma$.)
A: Actually, for free abelian groups something does in fact happen along the lines of what you ask. It is just that the semidirect product idea is a little bit of a red herring.
In the $Z^n$ case, what happens is that any virtually $Z^n$ group surjects onto one of the finitely many $n$-dimensional Euclidean crystallographic groups, with finite kernel. This is basically the content of the Bieberbach theorems. In the rank 2 case, that's 17 different Wallpaper groups to surject onto. In rank 3, that's 219 different Space groups.
A: It looks like the answer to your first question is no for both of the two classes of groups $F$ that you ask about, free groups and free abelian groups.
${\mathbb Z}_2 * {\mathbb Z}_3$ is virtually $F_2$ (free of rank 2), with quotient ${\mathbb Z}_6$, but it does not map onto a semidirect product of $F_2$ by a finite group, basically  because its finite subgroups have order at most 3.
The group $\langle x,y \mid y^{-1}xy=x^{-1} \rangle$  has a subgroup $\langle x,y^2 \rangle$ of index 2 isomorphic to ${\mathbb Z}^2$, but it does not surject onto a semidirect product of ${\mathbb Z}^2$ with a finite group.
A: Let me elaborate on the comment I made in the case of a virtually free group. Suppose $G$ is a virtually $F_n$ group, $F_n \leq  G$ a finite-index free group of rank $n$. By Stallings' theorem, $G$ is a graph of finite groups. In particular, there is a tree $\mathcal{T}$ and an action of $G$ on $\mathcal{T}$, such that $\mathcal{T}/G$ is a finite graph. The action of $F_n$ on $\mathcal{T}$ is faithful and free, since the stabilizer of any vertex in the $G$ action on $\mathcal{T}$ is finite. Consider the kernel $K=ker\{ G \to Aut(\mathcal{T})\}$. Then $K$ is finite since it stabilizes any vertex of $\mathcal{T}$. Then we should think of $\Lambda=\mathcal{T}/(G/K)$ as an orbispace, in the sense of  Haefliger, where we attach to each cell of $\Lambda$ the stabilizer in $G/K$ of a lift of the cell to $\mathcal{T}$. Also, $F_n\hookrightarrow G/K$, since $F_n \leq G$ acts faithfully on $\mathcal{T}$. The quotient $\mathcal{T}/F_n$ is a finite graph $\Gamma$ with $b_1(\Gamma)=n$ and no vertices of degree 1. There are only finitely many topological types of such graphs, and only finitely many orbispace covers $\Gamma \to \Lambda$. Thus, there are only finitely many possibilities for $G/K$. These groups are maybe the analogues of $\mathbb{Z},D_\infty$ in your question, or of the Bieberbach groups in Mosher's answer in the $\mathbb{Z}^n$ case.    
A: Let me add another class of examples: virtual hyperbolic surface groups, by which I mean groups that have some $\pi_1(S_g)$ as a finite index subgroup, where $S_g$ is the closed, oriented surface of genus $g \ge 2$. Each virtual hyperbolic surface group $\Gamma$ surjects, with finite kernel, onto the fundamental group of a closed hyperbolic 2-orbifold (predecessors to Haefliger's orbispaces mentioned above by Agol). And if you fix $g$ then there are only finite many such orbifold groups which are the targets of a virtual $\pi_1(S_g)$ group.
This class of examples comes from quasi-isometric rigidity of the class of hyperbolic surface groups, a theorem of Gabai and of Cannon and Jungreis, which says that any finitely generated group quasi-isometric to the hyperbolic plane (which includes virtual hyperbolic surface groups) has a surjection as described above for $\Gamma$.
One can mine the theory of quasi-isometric rigidity for other classes of examples.
