Suppose $P$ is a simple convex polyhedron with $n$ faces. Euler's formula and the handshaking lemma tell us that the number of edges $E=3n-6$. By a deformation of $P$, I mean a polyhedron, combinatorially equivalent to $P$, resulting from slightly perturbing the $n$ planes defining $P$. (Since $P$ is simple, any sufficiently small perturbation of faces preserves the combinatorial type.) Let's say two deformations are the same if there is a Euclidean isometry taking one to the other. A perturbation of each plane is given by 3 parameters (normal and position), and the Euclidean group is 6 dimensional. I believe this means that the space of deformations of $P$ is $(3n-6)$-dimensional—that is, $E$-dimensional. So, a deformation of $P$ is determined by one real parameter for each edge of $P$.
What could this parameter possibly be? Clearly not dihedral angle, as scaling is a non-trivial deformation which preserves all angles. Taking $P$ to be the prism of a rhombus, as suggested in this answer, demonstrates that it can't be edge length. But these two quantities are the only geometric properties of an edge that I can think of. Perhaps this is just a complete coincidence? Or perhaps my dimension argument is wrong.
