The standard name for this type of relation between two structures on $X$ is concordance rather than homotopy.  If two structures on $X$ are isotopic (with the respect to the appropriate homeomorphism group), then they are concordant, but not necessarily vice versa.  In some cases you can also assign a separate meaning to homotopy, but I don't think that it means the same thing as concordance.

There are then two levels to your question.  A triangulation $T$ of $X$ induces a piecewise linear structure $\mathcal{P}$.  You could ask whether the PL structures $\mathcal{P}$ and $\mathcal{P}'$ are isotopic or concordant, without worrying about the original triangulations.  For simplicity suppose that $X$ is a closed manifold.  Then at least in dimension $n \ge 5$, Kirby-Siebenmann theory says that the set of PL structures on $X$ up to isotopy are an affine space (or torsor) of $H^3(X,\mathbb{Z}/2)$.  I think that the concordance answer is the same, because the Kirby-Siebenmann invariant comes from the stable germ-theoretic tangent bundle of $X$.  In other words, two PL structures give you two different sections of a bundle on $X$ whose fiber is $\text{TOP}(n)/\text{PL}(n)$, the group of germs of homeomorphisms of $\mathbb{R}^n$ divided by the subgroup of germs of PL homeomorphisms.  Stabilization in this case means replacing $n$ by $\infty$ by adding extra factors of $\mathbb{R}$.  If $n = 4$, then up to isotopy there are lots of PL structures on many 4-manifolds, as established by gauge theory.  But I think that the concordance answer is once again Kirby-Siebenmann.  (I learned about this stuff in a seminar given by Rob Kirby  — I hope that I remembered it correctly!  You can also try [the reference by Kirby and Siebenmann][1], although it is not very easy to read.)

There is a coarser answer than the one that I just gave.  I tacitly assumed that the triangulations not only give you a PL structure (which always happens), but that they specifically give you a PL manifold structure, with the restriction that the link of every vertex is a PL sphere.  These are called "combinatorial triangulations".  It is a theorem of Edwards and Cannon that $S^5$ and other manifolds also have non-combinatorial triangulations.  If your question is about these, then it is known that they are described by some quotient of Kirby-Siebenmann theory, but it is not known how much you should quotient.  It is possible that every manifold of dimension $n \ge 5$ has a non-combinatorial triangulation, and that PL structures are always concordant in this weaker sense.  It is known that you should quotient more than trivially, that there are manifolds that have a non-combinatorial triangulation but no PL structure.  (I think.)

The other half of the question is to give $X$ a distinguished PL structure $\mathcal{P}$, and to look at triangulations $T$ and $T'$ that are both PL with respect to $\mathcal{P}$.  In this case there are two good sets of moves to convert $T$ to $T'$.  First, you can use stellar moves and their inverses.  A stellar move consists of adding a vertex $v$ to the interior of a simplex $\Delta$ (of some dimension) and supporting structure to turn the star of $\Delta$ into the star of $v$.  The theorem that these moves suffice is called the stellar subdivision theorem.  (The theorem is due to Alexander and Newman and it is explained pretty well in [the book by Rourke and Sanderson][2].)  The other set of moves are specific to manifolds and they are the bistellar moves or Pachner moves.  One definition is that a bistellar move is a stellar move that adds a vertex $v$ plus a different inverse stellar move that removes $v$ (hence the name).  But a clearer definition is that in dimension $n$, a bistellar move replaces $k$ simplices by $n-k+1$ simplices in the minimal way, given by a local $n+1$ concordance that consists of attaching a single $n+1$-dimensional simplex.  The theorem that these moves work is [due to Pachner][3].  Pachner's moves in particular give you a shellable triangulation of $X \times [0,1]$.

Even though the bistellar moves are motivated by concordance and the bistellar moves are not, it is not hard to make concordance triangulations for stellar moves as well.

  [1]: http://press.princeton.edu/titles/648.html
  [2]: http://www.worldcat.org/oclc/264145137
  [3]: http://portal.acm.org/citation.cfm?id=107898