When is a Lie group action is a local diffeomorphism?

Question

My whole question is actually in the title. Suppose a Lie group $G$ acting smoothly on a smooth manifold $M.$ I am looking for the conditions on the group action or the Lie group $G$ that makes $M$ and $g(M)$ are locally diffeomorphic for all $g\in G.$ Perhaps this is the case for all Lie group actions?

I was pondering about this question sometimes but couldn't get in to any conclusion. This might be obvious for the experts and people with sufficient background.

Answer 1

If I'm interpreting your (edited) question correctly, the answer is yes: For any element $g\in G$, $\theta_g:M\to M$ is smooth, since the action itself is smooth, and has a smooth inverse $\theta_{g^{-1}}:M\to M$, since $\theta_{g^{-1}}\circ\theta_g=\theta_{g^{-1}g}=\theta_e=\text{id}_M$. Thus each $\theta_g$ is a global diffeomorphism.

Answer 2

Possible only if $G$ is a discrete Lie group($0$-dimensional Lie group). A Lie group action on a manifold $M$ is first of all a smooth function $\theta: G \times M \to M$. If $\dim(G) >0$, then $\dim(G\times M) > \dim(M)$. So the the tangent spaces at each point can't be isomorphic(which is a necessary condition for a local diffeomorphism).

Now if $G$ is a discrete Lie group then the action can be a local diffeomorphism. For instance, if $G = \Bbb Z \times \Bbb Z$ and $M = \Bbb R^2$, then $G$ acts properly discontinuosly on $M$ via left translations and you can check that this action is a local diffeomorphism.