gt.geometric topology - Does a triangulation without fixed simplex property always exist?

EDITED. The arugment related to Mostov rigidity is completed according to a nice suggestion of Tom Church

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 non-identical 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 $phi$ does not collapse 3 simplexes of $M^3$ or identify them, then it is a homeomorphism of $M^3$ of finite order (Lemma above). From Mostov rigidity it follows that this automorphism is homotopic to the identity. In order to show that it IS in fact the identity we need to use a more involved statement suggseted by Tom Church below. Namely, a paritial case of Proposition 1.1 in http://www.math.uchicago.edu/~farb/papers/hidden.pdf
says that for a hyperbolic 3-manifold the group of isometries of any Riemanninan metric on it is isomorphic to a subgroup of the group of hyperbolic isometries. By our choice the group of hyperbolic isometries of $M^3$ is trivial. It is clear that $phi$ preserves a Riemannian metric on $M^3$. So by Prop 1.1 it is the identity.

From this it immediately follows that $phi$ sends $H^3(M^3, Z)$ to zero (since the volume is contacted). End of proof.