Classification of all equivariant structure on the Möbius line bundles Is there a classification of all equivariant structures of the Möbius line bundle $\ell\to S^1$?.
For example the antipodal action of $\mathbb{Z}/2\mathbb{Z}$ on $S^1$ cannot  be lifted to the total space $\ell$ to get an equivariant structure. But what about the general case?
In particular, let $\phi_{\theta}$ be the irrational rotation of the circle by $\theta$. Can the action of $\mathbb{Z}$ on $S^1$ given by  $n.x=\phi_{\theta}^n(x)$ be lifted to an action on the total space of the  Möbius bundle to give us an equivariant bundle? 
 A: Most of my work is with actions of compact Lie groups, so forgive me if I make any errors below related to other groups.
I'm not sure how general a classification you're looking for, but I'm going to assume a framework in which I can say something: Consider a group $G$ acting on the base space $S^1$ by rotations, so the action is given by a group homomorphism $G\to S^1$ and the usual action of $S^1$ on itself. Can that action be extended to an action on the Möbius bundle $\mathscr l$ by bundle maps?
Consider the space of unit vectors in $\mathscr l$. This space is a copy of $S^1$ and the projection to the base space is the double cover $\delta\colon S^1\to S^1$, $\delta(e^{i\theta}) = e^{2i\theta}$. We can also think of $\delta$ as the principle $\mathbb Z/2\mathbb Z$-bundle associated to $\mathscr l$.


Proposition If $G$ acts on the base space via $f\colon G\to S^1$, then the action extends to $\mathscr l$ if and only if $f$ lifts to a homomorphism $\bar f\colon G\to S^1$ such that $\delta\bar f = f$.


Proof: For clarity, write $B$ for the base space of $\mathscr l$ and $E$ for the set of unit vectors, so that each is a copy of $S^1$ and the projection $E\to B$ is the double cover $\delta$. If $G$ acts on $\mathscr l$, with its action on $B$ being by rotations, then it clearly acts on $E$ by rotations as well, which will be given by a homomorphism $\bar f\colon G\to E$ such that $\delta\bar f = f$. Conversely, given such an $\bar f$, it defines an action of $G$ on $E$ covering the action on $B$, and this extends to an action of $G$ on $\mathscr l$ by bundle maps. QED
So then you would want to determine what homomorphisms $f$ lift up the double cover. I don't know a general answer, but I suspect that, if you restrict to monomorphisms $f$, then the answer is that this is the case iff $G$ contains no element of order 2. This is certainly the case for finite subgroups of $S^1$, but there may be topological issues for infinite subgroups. (Or I could be missing something blindingly obvious, always a possibility.)
On the other hand, we can answer your last question: The homomorphism $f_\theta\colon \mathbb Z \to S^1$ given by $f_\theta(n) = e^{2\pi in\theta}$, with $\theta$ irrational, does lift to $\bar f_\theta$, given by $\bar f_\theta(n) = e^{\pi in\theta}$. In fact, there's no need to assume that $\theta$ is irrational, this works for any $\theta$, it's just that $f$ won't be a monomorphism unless $\theta$ is irrational, so in that case we can't think of this as lifting the action of a subgroup.
Note that I'm playing a bit fast and loose with topologies here. The proposition above can be interpreted as dealing with continuous $f$ and a continuous lifting/action. But thinking about topologies, you have to remember that, when $\theta$ is irrational, $f_\theta$ is a monomorphism but not a homeomorphism onto its image, so you have to be careful thinking of this as a lifting of the action of a subgroup.
