A topological space $X$ satisfies "Periodic orbit property", briefly POP, if for every continuous map

$f:X \to X$, there exist a natural number $n$ and a point $x_{0}\in X$ such that $f^{n}(x_{0})=x_{0}$

Obviously fixed point property(FPP) implies POP.

For a natural number $n$,a topological space $X$ is called $n-POP$ if for every continuous map $f$ on $X$, $f^{n}$ has a fixed point.(Ex: $\mathbb{S}^{2n}$ is a 2-POP manifold, using degree of maps)

The Question:

Is there an example of a manifold $M$ which satisfies POP but for every $n\in \mathbb{N}$, there is a continuous map $f$ on $M$ such that $f^{n}$ has no fixed point?

Namely: we search for a manifold for which every self map has a periodic orbit, but there is no any control on periods.

Equivalently:

Is there a manifold $M$ which is POP but not $n-POP$ for all $n\in \mathbb{N}$?

In particular, can we say:

"every compact POP manifold is necessarily a $n$-POP manifold, for some $n$"?

Motivating by Lefschetz fixed-point theorem, we ask that:

What algebraic topological criterion, can be introduced for consideration of this property(POP)?

**Edit:** According to the very interesting answer of Qiaochu Yuan, in the orientable case, the question is equivalent to the following:

Let M be a closed orientable manifold. Is it true that $M$ is not POP if and only if $\chi(M)=0$?

**Note:** I think the continuation of the argument of Qiaochu Yuan for his first statement is not easy, for arbitrary manifold. Because for the simplest case $S^{3}$ we had the famous conjecture of "existence of a vector field on $S^{3}$ without periodic orbit. In fact consideration of non vanishing vector fields is necessary but not sufficient. Periodic orbits of vector fields are important, too. Moreover, perhaps an approach which is not based on "vector fields" could be useful, for example consideration of oriention reversing diffeomorphisms.

notFPP? – Benjamin Dickman Jan 1 '14 at 12:38