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 $ <V,\tau> $ is exactly the stabilizer of the $ y $-axis.
The answer is yes.
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.
Both of these answers are copied directly from
https://mathoverflow.net/questions/414402/is-it-possible-to-realize-the-moebius-strip-as-a-linear-group-orbit