If we have a regular $n$-gon then we may consider the group of isometries: The dihedral group $D_{2n}$ generated by a reflection and a rotation.
Given a triangulation of an $n$-gon, these actions will produce new triangulations, but it may produce the same one. Consider the triangulation of the square given by joining the top left corner to the bottom right corner. A reflection along any axis will preserve this, but a rotation of $\pi/2$ radians will not.
My question is: Is it possible to have a triangulation of an $n$-gon that is rigid? That is, only the trivial isometry will preserve it.
Yes for $n \ge 7$. Consider the triangulation that uses all the diagonals from one vertex. That is preserved only by a reflection. In the first quadrilateral on one side use the other diagonal.
This picture when $n=6$ illustrates the principle, but fails since it's invariant under a rotation through $\pi$.
That suggests trying to settle the question for $5$ and $6$ as special cases.
(It might be reasonably easy to count the number of rigid triangulations. Perhaps the degree sequence of the vertices would help.)