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, 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.) 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. 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 stellar moves are not, it is not hard to make concordance triangulations for stellar moves as well.