Suppose we are given a triangulable topological space $X$. If $X$ has the fixed point property (FPP), then obviously for every triangulation $K$ of $X$ and every simplicial map $f:K\to K$ a simplex $\sigma\in K$ exists such that $f(\sigma)=\sigma$. This will be called the fixed simplex property (FSP).
One can give examples of simplicial complexes (included below) with FSP whose geometric realizations do not have FPP. Moreover, triangulations of spaces without FPP exist whose iterated barycentric subdivisions all have FSP.
Is it true that if $X$ is a triangulable space without FPP, then a triangulation of $X$ exists that does not have FSP? A related question: given a triangulation $K$ of $X$, can we find a (non-barycentric) subdivision of $K$ that does not have FSP?
Example (taken from J.A. Barmak's thesis (p. 101), who in turn cites K.Baclawski and A.Björner "Fixed points in partially ordered sets" (Adv. Math. 1979)):
Consider the regular CW-complex (square + 4 triangles) $C$ which is the border of a pyramid with square base. $C$ is homeomorphic to $S^2$, so it doesn't have FPP. Let $K$ be the following subdivision of $C$: divide each of the 3 triangular sides into 3 triangles by adding a vertex in the middle and divide the bottom square into 4 triangles the same way.
Name the top vertex of the pyramid $x$. Let $f:K\to K$ be simplicial. If $f$ is onto (on the vertices), then, by finiteness of $K$, $f$ is an automorphism. Since $x$ is the uniqe vertex that belongs to exactly eight 1-simplices, it is a fixed point. If $f$ is not onto, then $|f|$ is nullhomotopic and thus the Lefschetz number $\lambda(f)=1$. $|f|$ has a fixed point, so $f$ fixes a simplex.
The same argument (just count the 1-simplices that $x$ and other vertices belong to) applies to arbitrary barycentric subdivisions of $K$.
However, if in $C$ we subdivide only the square base into 4 triangles by adding a vertex in the middle, then the obtained simplicial complex doesn't have FSP.