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> Let $n>1$. Which smooth manifolds admit a diffeomorphism $f$ of order $n$?
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For a closed orientable surface $S\_g$ of genus $g$ and $n=2$ the answer is in the affirmative, since $S\_g$ can be embedded in $\mathbb{R}^3$ in a symmetric way. A similar argument gives a positive answer for $n=3$: $S\_g$ can be embedded in $\mathbb{R}^3$ with rotational symmetry of order 3. The first problem I've encountered in case of $S\_g$ is with $g=2$ and $n=4$: all my candidates for $f$ turned out to have order $2$. And I have no idea as to what happens for $g=2$ and $n=5$.


In the case of a general manifold $M$, the existence of a nontrivial global flow $\phi\_t$ such that $\phi\_0=\phi\_1$ would give a positive answer for every $n$, but it seems to be a much stronger condition.


Is a general answer to this question known? If not, maybe there are some simple examples of manifolds without diffeomorphisms of given order?



Which manifolds admit a diffeomorphism of order $n$?
The Nielsen Realisation Problem asks when a (finite) subgroup of the mapping class group (the group of isotopy classes of diffeomorphisms) of a surface can be realised as a group of diffeomorphisms. Kerckhoff proved in the 80s that **every** finite subgroup of the mapping class group can be realised. (For infinite subgroups, there are various known obstructions, such as the Miller-Morita-Mumford characteristic classes.) Thus, Kerckhoff's theorem implies that a surface admits a diffeomorphism of order n if its mapping class group has an element of order n. Conversely, one can show that any diffeomorphism of a surface must have infinite order if it is isotopic to the identity, every diffeomorphism of order n gives an order n mapping class group element.


If you have a diffeomorphism of finite order on a surface then you can find a complex structure (or a Riemannian metric or a symplectic structure or a conformal structure) for which the diffeomorphism is an automorphism/isometry. This is accomplished by choosing an arbitrary metric and then averaging over all translates of it by powers of the diffeomorphism.


So the point of all this is that in dimension 2 finding an order n diffeomorphism of a genus g surface is the same as finding a complex curve with an order n isometry, or equivalently, a Z/n orbifold point in the moduli space. As Sam and Robin alluded to, there is a bound on the order of n relative to g. Hurwitz's theorem states that the order of the automorphism group of a genus g curve is less than or equal to 84(g−1). There are various other theorems that tell you about what sorts of finite subgroups you can find in mapping class groups.


In higher dimensions, it's harder to give a useful answer. If your manifold has a circle action then you are done because $S^1$ contains Z/n for any n. But there are plenty of manifolds around which do not admit circle actions, such as K3 surfaces. The $\widehat{A}$-genus is an obstruction to admitting a circle action. Some nice things are known about the finite groups of automorphisms of K3 surfaces. In fact, I think they are pretty much completely classified into a finite list.


In general, by averaging over translates of a metric, you can still assume that a given finite order diffeomorphism acts by isometries for some metric. Generally, the isometry group of a compact Riemannian manifold will be a finite dimensional compact Lie group (I think this is a theorem of Yau). 


