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 $t_{0}>0$ be small enough so that $\varphi(t_0)$ has no fixed points. Such $t_{0}$ exists, because there is a positive uniform lower bound for the period of all periodic orbits.(As a consequence of the flow box theorem, around regular points of a vector field). 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. $\Box$
Best Answer
Lovely question! Sadly, the answer is "no" in the sense that the fixed point property is not homotopy-invariant even in the category of finite polyhedra. In fact, it is also not invariant under the operations of taking products or suspensions.
See the three page paper of W Lopez called "An example in the fixed point theory of polyhedra" for the construction of an explicit counterexample to your desired property as well as the two properties listed above. Basically, Lopez's construction involves two finite polyhedra $X$ and $Y$ whose wedge product has the fixed point property but whose union along an edge does not (!!). The Corollary to Theorem 3 on the second page of the linked pdf is of interest.
Update (4th Oct 2015): I have also been looking for positive results lately, and one good source is Robert Brown's Handbook of Topological Fixed Point Theory (the Google book is here. Theorem 8.11 in the book is this cool result of Jerrard:
So if you can decompose your space as a nice enough product of fixed point spaces, then there is some hope depending on their homology...