On MSE this got 5 upvotes but no answers not even a comment so I figured it was time to cross-post it on MO:

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

**Does there exists a Lie group $ G $ a representation $ \pi: G \to \operatorname{Aut}(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 $ \operatorname{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 $$ \operatorname{SE}_2= \left \{ \ \begin{bmatrix} a & b & x \\ -b & a & y \\ 0 & 0 & 1 \end{bmatrix} : a^2+b^2=1 \right \}. $$ 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 $$ \operatorname{SE}_2/\langle V, \tau \rangle $$ is the Moebius strip.

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