Nice question! Here's what I can show.

Let $X$ be a smooth closed manifold. Then:

(1) If $\chi(X) = 0$, then $X$ is not $n$-POP for any $n$.

(2) If $\chi(X) \neq 0$ and $X$ is orientable, then $X$ is $\text{lcm}(1, 2, ... n)$-POP with respect to maps $f : X \to X$ of nonzero degree, where $n = \text{max}(b_0 + b_2 + ..., b_1 + b_3 + ...)$ (where $b_i$ is the $i^{th}$ Betti number of $X$).

*Proof of 1.* We will use the converse of the Poincaré-Hopf theorem: if $\chi(X) = 0$, then $X$ admits a nonvanishing vector field. Let $\varphi(t)$ denote the flow of this vector field. Let $t_0$ be small enough so that $\varphi(t_0)$ has no fixed points. For a given $n \in \mathbb{N}$, let $f = \varphi \left( \frac{t_0}{n} \right)$. Then $f^n$ has no fixed points, hence $X$ is not $n$-POP. $\Box$

(I strongly suspect that in this case $X$ is not POP either; it seems like we should be able to consider a small flow of a sufficiently generic nonvanishing vector field. But I don't know how to finish this argument.)

*Proof of 2.* We will need the following two observations.

**Lemma 1:** Let $f_0, f_1$ be linear operators acting on two finite-dimensional vector spaces $V_0, V_1$. If $\text{tr}(f_0^k) = \text{tr}(f_1^k)$ for $k$ between $1$ and $\text{max}(\dim V_0, \dim V_1)$, then $f_0$ and $f_1$ have the same nonzero eigenvalues with the same multiplicities.

*Proof.* The above condition implies, using the Newton-Girard identities, that $f_0$ and $f_1$ have the same characteristic polynomial up to factors of $t$. $\Box$

**Lemma 2:** Let $X$ be an $n$-dimensional smooth closed oriented manifold and let $f : X \to X$ be a map of nonzero degree. Then every eigenvalue of $f$ acting on cohomology (with complex coefficients) is nonzero.

*Proof.* Let $e_1, ..., e_d$ be a basis of generalized eigenvectors for the action of $f$ on $H^k(X, \mathbb{C})$. By Poincaré duality the cup product $H^k \otimes H^{n-k} \to H^n$ is nondegenerate, so we can find a dual basis $e_1^{\ast}, ..., e_d^{\ast}$ of $H^{n-k}(X, \mathbb{C})$. Since $f$ acts by a nonzero scalar, namely $\deg f$, on $e_i \smile e_i^{\ast}$ for all $i$, the generalized eigenvalue of $e_i$ must also be nonzero. $\Box$

Now back to the proof of 2. With hypotheses as above, let $f_0$ denote the map induced by $f$ on the direct sum $V_0$ of the even-dimensional complex cohomology of $X$ and let $f_1$ denote the map induced by $f$ on the direct sum $V_1$ of the odd-dimensional complex cohomology of $X$, so that the Lefschetz trace of $f^k$ can be written

$$L(f^k) = \text{tr}(f_0^k) - \text{tr}(f_1^k).$$

By Lemma 2, the eigenvalues of $f_0$ and $f_1$ are all nonzero, so if $f_0$ and $f_1$ have the same nonzero eigenvalues then in particular $\dim V_0 = \dim V_1$. By the contrapositive of Lemma 1, if $\chi(X) = \dim V_0 - \dim V_1 \neq 0$, then there exists some $k$ between $1$ and $n = \text{max}(\dim V_0, \dim V_1)$ such that $L(f^k) \neq 0$, hence, by the Lefschetz fixed point theorem, such that $f^k$ has a fixed point. In particular, $f^{\text{lcm}(1, 2, ... n)}$ has a fixed point. $\Box$

notFPP? – Benjamin Dickman Jan 1 at 12:38