How bad can $\pi_1$ of a linear group orbit be? Let $G$ be a simply connected Lie group and $\mathcal O= G(v)=G/G_v$ a $G$-orbit in some finite-dimensional $G$-module $V$. By the homotopy exact sequence, its fundamental group $\Gamma$ is the component group of the stabilizer $G_v$:
$$
\Gamma= \pi_1(\mathcal O) = \pi_0(G_v) = G_v/G_v^{\mathrm o}.
$$

Can one characterize the class of discrete groups $\Gamma$ that occur in this way (for some orbit in some representation of some $G$)?
  Failing that, what's the most complicated $\Gamma$ you've seen occur?

My (very possibly misguided) hunch is that e.g. $\mathbf{SL}(2,\mathbf Z)$, or the discrete Heisenberg group, don't occur "because I would have heard of it". On the other hand, it is clear that:
1. Every finite group $\Gamma$ occurs.
Indeed, using e.g. its regular representation, we can embed $\Gamma$ as a closed subgroup of $G =\mathbf{SU}(N)$ for some $N$; and then the Palais-Mostow theorem guarantees [e.g. Bourbaki, Lie Groups, IX.9.2, Cor. 2] that $\Gamma$ is the stabilizer of some vector in some $G$-module, q.e.d. 
2. $\mathbf Z^n$ occurs.
Indeed, letting the additive group $G=\mathbf R^n$ act on $V=\mathbf C^n$ by
$
\smash{\begin{pmatrix}
e^{2\pi ig_1}\\
&\ddots\\
&&e^{2\pi ig_n}
\end{pmatrix}}
$
it is clear that $v={}^t(1,\dots,1)$ has stabilizer $\mathbf Z^n$.
 A: I can show that the group is virtually abelian.
Let $L$ be the Zariski closure of the image of $G$ and let the stabilizer of the point be $L_v$. Then, as YCor points out, the image of $G$ is a normal subgroup of $L$.
We may also force it to be a closed subgroup. We do this by taking the direct sum of our representation with a representation of $G/ [G,G]$ in which it embeds as a closed subgroup of $GL_n$, which we can do because $G$ is a simply connected Lie group, hence $G/[G,G]$ is $\mathbb R^n$. Because the image of $G$ in the product of two representations, one faithful with a closed embedding, is the graph of the map to the other representation, and the graph is closed by the closed graph theorem, it is now closed.
Because it is closed and normal, we may take a quotient by it. Thus the algebraic orbit $L/L_v$ maps to $L/j(G) L_v$ and the fiber is $j(G) / j(G) \cap L_v= \mathcal O$. Hence we have an exact sequence:
$$\pi_2( L/ j(G) L_v) \to \pi_1( \mathcal O) \to \pi_1(L/L_v) $$
Because, as YCor points out, $j(G)$ contains the commutator subgroup of the identity component of $L$, $L/j(G) L_v$ is a union of abelian Lie groups, hence its universal cover is $\mathbb R^n$, so its $\pi_1$ vanishes. Hence $\pi_1(\mathcal O)$ is contained in $\pi_1(L/L_v)$.
Now we have an exact sequence:
$$\pi_1(L) \to \pi_1(L/L_v) \to \pi_0(L_v)$$
$\pi_1(L)$ is abelian, because $L$ is a Lie group, and $\pi_0(L_v)$ is finite, because $L_v$ is an algebraic group, so $\pi_1(L/L_v)$ is virtually abelian, hence $\pi_1(\mathcal O)$ is virtually abelian. $\pi_1(L)$ is also finitely generated, so $\pi_1(\mathcal O)$ is finitely generated.
In fact, $\pi_0(L_v)$ acts trivially by conjugation on $\pi_1(L)$, so we have a central extension of a finite group by an finitely generated abelian group. By changing the subgroup, we can make this a central extension of a finite group by $\mathbb Z^d$. 
A: Here is a geometric proof which shows that the fundamental group of a linear group orbit must be finite by abelian (confirming the most general form of the two answers given above).
By theorem of Mostow mentioned in this question
Homogeneous manifold deformation retracts onto compact submanifold
["Covariant Fiberings of Klein spaces" Mostow 1955]
If G and G' both have finitely many connected components (for example if they are algebraic groups) then G/G' is a vector bundle over K/K' where K and K' are maximal compacts.
This is the case here because the image of a representation is always an algebraic group and the stabilizer of a vector is always Zariski closed and thus always an algebraic group.
The total space of a vector bundle is homotopy equivalent to the base. Thus G/G' is homotopy equivalent to K/K'.
Now using the fact that K has a universal cover which is a product of a compact and an abelian factor we can argue as in
https://math.stackexchange.com/questions/4321106/transitive-action-by-compact-lie-group-implies-almost-abelian-fundamental-group?noredirect=1#comment9009256_4321106
that the fundamental group of K/K' , and thus of G/G', is indeed finite by abelian
A: I think we only obtain central-by-finite-by-abelian finitely generated groups (I don't claim we get all of them; maybe we only get finite-by-abelian groups and this is what I expect).
(I use the convention that a group is P-by-Q if it lies in an exact sequence with kernel satisfying P and quotient satisfying Q.)
This is enough to discard $\mathrm{SL}_2(\mathbf{Z})$, torsion-free nilpotent groups of nilpotency class $\ge 3$, or even many virtually abelian groups such as the infinite dihedral group, but not the discrete Heisenberg group.
Actually, a central-by-(finite-by-abelian) f.g. group is either finite-by-abelian or contains a copy of the discrete Heisenberg group so in a sense it's the last guy to rule out (if what follows is correct).

Let me assume only that $G$ is a connected Lie group because simple connectedness does not seem to play any role. Clearly the question does not change if we mod out by the unit component of the kernel of the representation, so we can suppose that the representation has a discrete (hence central) kernel.
Let $L_a$ be the Zariski closure of the image of the representation $j:G\to\mathrm{GL}(V)$; hence $L_a$ is Zariski connected. Write $H_a=(L_a)_v$ (the stabilizer of $v$), so $H_a$ is Zariski-closed in $L_a$. Then $G_v=j^{-1}(H_a)$.
So now we have: a connected Lie group $G$, a continuous homomorphism $j$ with discrete kernel and Zariski-dense image into a real connected linear algebraic group $L_a$, a Zariski-closed subgroup $H_a$ in $L_a$, and we wish to understand the group of components of $j^{-1}(H_a)$. The group of real points of $L_a$ has finitely many components in the Lie topology; let $L$ be the unit component, and let $H=L\cap H_a$. Since $j(G)\subset L$, we have $j^{-1}(H)=j^{-1}(H_a)$.
Since $j(G)$ is Zariski dense in $L$, its Lie algebra is an ideal in the Lie algebra of $L$ with abelian quotient. In particular (by Zariski connectedness of $L$) its Lie algebra is normalized by $L$, and thus by connectedness of $G$, we obtain that $j(G)$ is normal in $L$ with abelian quotient $(*)$. In particular, the derived subgroup $L'$ of $L$ (viewed as the group of real points) is contained in $j(G)$. (This derived subgroup has finite index in the group of real points of the algebraic group $[L_a,L_a]$.)
Let $p$ be the projection $L\to L/L'$. Then $L/L'$ is an abelian connected Lie group; $p(H)$ is closed in $L/L'$; the image $p\circ j(G)\subset L/L'$ is a connected immersed Lie subgroup, and hence $(p\circ j)^{-1}(p(H))/\mathrm{Ker}(p\circ j)$, as a closed subgroup of a connected abelian Lie group, is isomorphic to the direct product of a connected abelian Lie group and a free abelian group of finite rank.
By definition, we have $\mathrm{Ker}(p\circ j)\cap j^{-1}(H)=j^{-1}(L'\cap H)$, which equals $j^{-1}(L')\cap j^{-1}(H)$. Recall that $L'\cap H\subset L'\subset j(G)$, so its Lie algebra is the image by $j$ of some Lie subalgebra of the Lie algebra of $G$, generating an immersed subgroup $M$, which is normal in $j^{-1}(H)$. Then $j(M)$ is the unit component of $L'\cap H$ (in the Lie topology), so $\mathrm{Ker}(j)M$ has finite index in $j^{-1}(L'\cap H)$.  
To summarize, we have the inclusions $\mathrm{Ker}(j)M\subset j^{-1}(L'\cap H)\subset j(H)$; these are closed normal subgroups, $M$ is connected, $\mathrm{Ker}(j)$ is central, the quotient $j^{-1}(L'\cap H)/\mathrm{Ker}(j)M$ is finite and the quotient $j^{-1}(H)/j^{-1}(L'\cap H)$ is abelian. Let $p$ be the projection $j^{-1}(H)\to \Gamma=\pi_0(j^{-1}(H))$. Then $p(M)$ is trivial, so $p(\mathrm{Ker}(j)M)=p(\mathrm{Ker}(j))$ is central; the quotient $p(j^{-1}(L'\cap H))/p(\mathrm{Ker}(j)M)$ is finite and the quotient $p(j^{-1}(H))/p(j^{-1}(L'\cap H))$ is abelian. Thus $\Gamma$ is central-by-(finite-by-abelian). [Since central-by-finite implies finite-by-abelian, $\Gamma$ is also central-by-(2-step-nilpotent)].

Proof of the claim that the Lie algebra $\mathfrak{p}$ of $j(G)$ is a ideal in the Lie algebra $\mathfrak{l}$ of $L$ and $\mathfrak{l}/\mathfrak{p}$ is abelian:

First, $j(G)$ normalizes $\mathfrak{p}$; since the normalizer in $L$ of $\mathfrak{p}$ is Zariski-closed and contains the Zariski-dense subgroup $j(G)$, it is equal to $L$. So $\mathfrak{p}$ is an ideal. Next, we use a theorem due to Chevalley (see e.g., Chevalley, Théorie des groupes de Lie, Hermann; Chap. II.14), which says that for any Lie subalgebra of $\mathfrak{l}$, its derived subalgebra is the Lie algebra of a Zariski-closed subgroup; thus let $Q$ be a Zariski-closed subgroup whose Lie algebra is $[\mathfrak{p},\mathfrak{p}]$. Then it follows that the derived subgroup of $[j(G),j(G)]$ is equal to the unit component of $Q$ (in the Lie topology). In particular, $[g,h]\in Q$ for any $g,h\in j(G)$. By Zariski density, it follows that $[L,L]\subset Q$. Thus $[\mathfrak{l},\mathfrak{l}]=[\mathfrak{p},\mathfrak{p}]\subset\mathfrak{p}$.

Update. I here answer the question in the comment: yes, every f.g. finite-by-abelian group occurs. Every such group $P$ can be written as fibered product $P=F\times_\Lambda\Gamma$, where $\Gamma$ is a free abelian group of finite rank, $F$ is a finite group, and $\Lambda$ is a common quotient (thus a finite abelian group). Choose $n$ and an embedding $F\to S=\mathrm{SL}_n(\mathbf{C})$, and an embedding of $\Lambda$ into a complex torus $T$ (defining by composition a map $F\to T$). Now embed $F$ diagonally into $S\times T$. Then the fundamental group of $(S\times T)/F$ is isomorphic to $P$ (see Corollary 3.1 of this note of mine with Borovoi; possibly there are earlier references). Finally use the theorem you refer to so as to realize $F$ as stabilizer of some vector for some $(S\times T)$-module.
