Suppose that $ M $ is non-orientable with transitive action by a Lie group $ G $. Does that imply that some Lie group $ G' $ acts transitively on the orientable double cover $M'$?
This is true for compact dimension 2: the Klein bottle is an $ \operatorname{SE}_2 $ manifold and so is the torus. The projective plane is an $ \operatorname{SO}_3 $ manifold and so is the sphere.
Morally I think the answer should be yes in general since an orientable double cover should be better behaved than/at least as well behaved as the original non orientable manifold.
Best Answer
There is a general theory for lifting Lie group actions to covering spaces, see Bredon's monograph "Introduction to compact transformation groups", chapter 1, section 9. In the case of orientation covers one gets that any (effective, continuous) action of a Lie group $G$ on a non-orientable manifold lifts to a $G^\prime$ action on the orientation cover, where $G^\prime$ consists of all lifts of elements of $G$. Thus $G^\prime$ surjects onto $G$ with kernel of order two, and in particular, $G^\prime$ is a Lie group.
If the $G$-action is transitive, then so is the $G^\prime$-action. (Indeed, if $\tilde x, \tilde y$ are points in the covering space, which project to $x, y$, respectively, and $g\in G$ with $g(x)=y$, then one of the two lifts of $g$ takes $\tilde x$ to $\tilde y$).