Note: after not receiving any answer for some time, I asked this in mathoverflow, and got an answer there.
The Question:
Is it possible for a diffeomorphism $\phi$ (of a smooth manifold $M$) to have the following properties:
- All its orbits are finite.
- $\phi$ is not of finite order. (and hence has arbitrarily large finite orbits).
There are such examples if we allow infinitely many connected components (Take for instance countable disjoint copies of $\mathbb{S}^1$ and rotation of order $n$ in the $n$-th copy).
So let us assume $M$ is connected. (I prove below that we can reduce the case of finitely many components to one component).
I will add that this question was inspired by a previous question of mine, about global obstructions for a diffeomorphism to be an isometry (You can see the update there for the relevant connection).
Update: It turns out the answer is negative. There is no such a diffeomorphism. (For details see the answer in mathoverflow).
Claim: existence of a self-diffeomorphism for a manifold with finitely many connected components, implies existence for a connected manifold.
Proof: In this case the components $U_i$ are clopen connected sets, hence $\phi(U_i)$ are also clopen connected sets. Now combine the following two facts:
- clopen sets are always a union of connected components.
- every connected subset of a topological space $X$ is contained in a connected component $X$.
Now it is evident that each $\phi(U_i)$ is a connected component, i.e $\phi$ permutes the components. Since there are finitely many components, if we choose some component $U_i$, we will get $\phi^n(U_i)=U_i$ for sufficiently large $n$.
Next, we observe that the if $\phi$ has the two required properties, then $\phi^n$ also has them.
So finally, $\phi^n|_{U_i}$ is a self-diffeomoprhism of a connected manifold with the required properties.
This has been successfully asked and answered on MO. The answer is no.