The answer to the first question is no. There exsit manifolds of dimension 3 such that every simlicial map of the manfiold to itself (for any simplicial decomposition) has a fixed point (and hence a fixed simplex). At the same time every 3-mafiold adimits a smooth self-map without fixed points.
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 take such $M^3$ that does not have isometries. Existence of such manifolds is a standard result of 3-dimensional hyperbolic geometry. Let us prove that every such manifold gives us 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 $M^3$. First we state a simple lemma (without a a proof)
Lemma. Consider a simplicial decomposition of a compact orienable manifold. Suppose we have a simpicial map from it to itself, that send simplexes of highest dimensions to simplexes of higher dimentions (i.e. don't collapse them) and don't indentify them. Then this is an automorphism of finite order.
Corollary. Every simplicial map $\phi$ from $M^3$ to itself either collapses a simplex of dimension 3 or identifies two such simplexes. In paricular the generator of $H^3(M^3,\mathbb Z)$ is sent to zero by this map.
This corollary 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 corollary. If $\phi$ would not collapse 3 simplexes of $M^3$ or identify them, it would be a homeomorphism of $M^3$ of finite order. 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.
From this it immediately follows that $\phi$ sends $H^3(M^3, Z)$ to zero (since the volume is contacted). End of proof.