Diffeomorphism group and covering spaces

Group actions on $M$ fail to lift to $N$ even for one of the most fundamental examples, namely the universal covering map $f : \mathbb{R} \mapsto S^1$ given by the formula $$f(t)=e^{2\pi \, t \, i} $$ For example, the group of rotations of $S^1$ does not lift to $\text{Diff}(\mathbb{R})$. This is true even though every individual rotation does lift, the trouble being that the lifts of elements are not unique, and cannot be chosen so as to preserve the group operations.

To be specific, letting $\rho : S^1 \to S^1$ be the rotation through angle $\pi$, it follows that $\rho^2 = \rho \circ \rho=\text{Id}$. But the lifts of $\rho$ are precisely the translations of $\mathbb{R}$ of the form $r(t) = t + (2k+1)\pi$, $k \in \mathbb{Z}$, and so $$r^2(t) = r \circ r(t) = t + (4k+2)\pi $$ which is not the identity, no matter what the value of $k$ is.

Just to be a bit contrary, group actions in sense do lift. Of course, Lee is right that if $G$ acts on $M$ nontrivally, there is no reason to assume $G$ acts on $N$ non-trivially (or vice versa). But you were allowed to replace $M$ with $N$ so it's only fair that you allow $G$ to be replaced as well.

For ease, I will assume $N\rightarrow M$ is the universal covering. Suppose a topological group $G$ acts on $M$.

This is defined by some kind of map $G\times M\rightarrow M$ satisfying certain conditions. We are looking for a map $H\times N\rightarrow N$ where $H$ is a topological group which could, in some sense, be called a lift of the $G$ action on $M$. To that end, let $H$ denote the universal cover of $G$.

Then we have a canonical map $H\times N\rightarrow G\times M\rightarrow M$. Since $H$ and $N$ are simply connected, the lifting criteria for coverings guarantees we can lift the map. If we fix $n\in N$, we know we want $(e,n)$ (where $e\in H$ is the identity) to map to $n$. If we specifiy this, then we get a unique lift $H\times N\rightarrow N$.

Now one can verify that this lift does in fact define a group action of $H$ on $N$. Further, if $\pi:H\rightarrow G$ and $\rho:N\rightarrow M$ are the universal covering maps, then for any $h\in H$, $n\in N$, one can verify that $\rho(h\ast n) = \pi(h)\ast\rho(n)$, so the actions are compatible in this sense.