Does there exists a Lie group $ G $, a finite dimensional representation $ \pi: G \to GL(V) $, and a vector $ v \in V $ such that the orbit
$$
\mathcal{O}_v=\{ \pi(g)v: g\in G \}
$$
is diffeomorphic to the Klein bottle?
The Klein bottle, $ K $, is homogeneous for the special Euclidean group of the plane
$$
SE_2= \left \{ \
\begin{bmatrix}
a & b & x \\
-b & a & y \\
0 & 0 & 1
\end{bmatrix} : a^2+b^2=1 \right \}
$$
Because there is a one parameter group $ H $ of translations up each vertical line
$$
H= \left \{ \
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & y \\
0 & 0 & 1
\end{bmatrix} : y \in \mathbb{R} \right \}
$$
There is also an element that shifts horizontally by one unit
$$
b:=\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
$$
If we mod out $ SE_2 $ by the group generated by $ \langle H,b \rangle \cong \mathbb{R} \times \mathbb{Z} $ we just get a torus. Now if we include the rotation by 180 degrees
$$
r:=\begin{bmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{bmatrix}
$$
Then mod out $ SE_2 $ by the group generated by $ \langle H,r,b \rangle $ we get the Klein bottle $ K $. Since $ \mathbb{Z}=\pi_1(SE_2) $ and $ \pi_0(<H,r,b>) =<r,b> \cong \{ \pm 1 \} \rtimes \mathbb{Z} $ then we must have
$$
0 \to \mathbb{Z} \to \pi(K) \to \mathbb{Z} \rtimes \{ \pm 1 \} \to 0
$$
So we can conclude that $ \pi_1(K) \cong \mathbb{Z} \rtimes (\mathbb{Z} \rtimes \{ \pm 1 \}) $. It turns out that this particular semidirect product is isomorphic to the simpler semidirect product $ \mathbb{Z} \rtimes \mathbb{Z} $ with presentation $ <a,b:abab^{-1}> $ . Thus $ \pi_1(K) \cong \mathbb{Z} \rtimes \mathbb{Z} $ (this fact can be verified using the Seifert-Van Kampen theorem and the fact that a Klein bottle is a connected sum of two projective planes)
As Steve D pointed out below, since the klein bottle has orientable double cover the torus then by applying LES of homotopy to the $C_2$ principal bundle we have
$$
1 \to \mathbb{Z}^2 \to \pi_1(KB) \to C_2 \to 1
$$
So the fundamental group is virtually abelian.
Best Answer
The answer is no.
A compact linear group orbit must have a transitive action by a compact Lie group (which is equivalent to being a compact manifold which admits a metric with respect to which it is Riemannian homogeneous) see
https://mathoverflow.net/questions/409511/compact-linear-group-orbit-equivalent-to-linear-compact-group-orbit
The klein bottle is compact and aspherical (universal cover is the plane). The only compact aspherical linear group orbit is the torus see
Riemannian homogeneous aspherical iff flat torus