Let me give an example of a compact manifold of dimension 3 that has no FPP, but has FSP.

Namely, take $M^3$ with vanishing first and second homology ($H_1(M^3,R)=H_2(M^3,R)=0$) and that is a hyperbolic 3-manifold.  Moreover we take such $M^3$ that does not have isometries. Such a manifold exists by a general theory. Let us prove that this gives an example.

Proof. All compact 3-manifolds have zero Euler characteristics, so on $M^3$ there is a non-vanishing vector field $v$. Take the flow $F_t$ generated by $v$ in small time $t$. This will give us a family of diffeo $F_t$ of $M^3$ that don't have fixed points for small $t$. So $M^3$ in not FPP. 

Now, let us show that $M^3$ has FSP. Take any simplicial decomposition of $M^3$. It introduces a singular polyhedral metric on $M^3$. Namely, we identify every $3$-simplex with a Euclidean simplex whose sides have length 1. 

Lemma. Every simplicial map $\phi$ from $M^3$ to itself is not an isometry and the generator of $H^3(M^3,\mathbb Z)$ is sent to zero by this map. 

This lemma together with Lefshetz fixed point theorem implies immediately that $\phi$
has a fixed point, and so it proves FSP for $M^3$ (we use that $H_1(M^3)=H_2(M^3)=0$).

Proof of Lemma. First of all $\phi$ can not be an isometry. Otherwise this would be a diffeo of $M^3$ of finite order, and by Mostov rigidity this would imply that the hyperbolic metric of $M^3$ has automorhisms. But by our assumption $M^3$ does not allow hyperbolic isometry.  

It follows, that the map is contracting for the polyhedral metric. Indeed, $\phi$ sends every edge to an edge or a point, so it can not increase the distances. Since it can not be an isometry, it should either collapse one of $3$-simplexes to a $2$-simplex (or an edge, or a point), or reverse the orientation of one of $3$ symplexes.
From this it immediately follows that $\phi$ sends $H^3(M^3, Z)$ to zero (since the volume is contacted). End of proof.