Suppose that a Lie group $ G $ acts transitively on a manifold $ M $. If $ G $ is compact then we can equip $ M $ with a Riemannian metric with respect to which $ G $ acts by isometries.
What if $ G $ is noncompact? Can we equip $ M $ with a Riemannian metric with respect to which $ G $ acts by conformal transformations?
The first counter example that comes to mind is take $ G=E_2 $ the isometry group of the plane. Then take $ M $ to be the Moebius strip or the Klein bottle. $ G $ acts transitively on both these spaces. But I have no idea if the action is conformal (say with respect to the flat metric).
Best Answer
No. Take $G=\mathrm{SL}(n,\mathbb{R})$ for $n \ge 3$ and consider the action on $\mathbb{R}\mathbb{P}^{n-1}$. If the action preserved the conformal class of a Riemannian metric $\langle \cdot,\cdot\rangle$, then the stabilizer $H$ of a point $[v]$ would preserve the conformal class of the metric on the tangent space $T_{[v]}\mathbb{R}\mathbb{P}^{n-1}$. But every linear transformation of $T_{[v]}\mathbb{R}\mathbb{P}^{n-1}$ is induced by an element of $H$, so $H$ cannot preserve a conformal class when $n \ge 3$.
More generally, you can find lots of actions where the point-stabilizer is "too big" to respect a conformal class.