Is it possible to realize the Moebius strip as a linear group orbit

algebraic-groupsdifferential-geometrygeometric-topologylie-groupsrepresentation-theory

Is the Moebius strip a linear group orbit? In other words:

Does there exists a Lie group $$G$$, a 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 Moebius strip?

My thoughts so far:

The only two obstructions I know for being a linear group orbit is that the manifold (1) must be smooth homogeneous (shown below for the the group $$SE_2$$) and (2) must be a vector bundle over a compact Riemannian homogeneous manifold (here the base is the circle $$S^1$$).

The Moebius strip is homogeneous for the special Euclidean group of the plane
$$SE_2= \left \{ \ \begin{bmatrix} a & b & x \\ -b & a & y \\ 0 & 0 & 1 \end{bmatrix} : a^2+b^2=1 \right \}$$
there is a connected group $$V$$ of translations up each vertical line
$$V= \left \{ \ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & y \\ 0 & 0 & 1 \end{bmatrix} : y \in \mathbb{R} \right \}$$
Now if we include the rotation by 180 degrees
$$\tau:=\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
Then $$\langle V, \tau \rangle$$ has two connected components and
$$SE_2 \mathbin{/} \langle V, \tau \rangle$$
is the Moebius strip. Indeed if we consider the model of the Moebius strip as the manifold of affine lines in the plane then the subgroup $$$$ is exactly the stabilizer of the $$y$$-axis.

Take $$O_{2,1}$$ and act on the second symmetric power of the standard rep $$\mathbb{R}^3$$ and the orbit of $$e_1 \otimes e_1$$ works where $$e_1,e_2,e_3$$ is standard basis.
Or take natural $$E_2$$ action of affine transformations on 6d space of all polynomials in $$x,y$$ of degree 2 or less. Then the orbit of $$x^2$$ works.