Given two polygons of equal area with horizontal foliations, can one describe the obstruction (if there is any but I suspect the answer to be yes) to scissor-equivalence respecting the horizontal foliation (only half-turns and translations allowed during scissor-equivalence)?

A (fairly easy) positive answer can be given if both polygons (of equal area, of course) have rational vertices (with respect to orthogonal coordinates). It is also easy to construct scissor-equivalent examples not satisfying (obvious generalizations of) such rationality conditions.

More generally, it is in fact easy to transform every foliated polygon into a union of (horizontally foliated) rectangles by scissor equivalence. I am however stuck when trying to transform two such foliated rectangles by scissor-equivalence respecting the folation.

One idea for showing that two such foliated rectangles with irrational height-rations are not "foliatedly scissor-equivalent" is to look at the permutation induced on the four vertical edges (minus a finite number of points) induced by such a scissor equivalence. Start at a point of a vertical edge, go horizontally until the first cut where you jump to the glued point on the second rectangle and continue going (perhaps backwards and jumping between the two rectangles at each cut) until you hit the boundary. This defines an involution which is piecewise isometric (an exchange of intervals ofall four vertical edges) with a finite number of discontinuities. I guess the obstruction (for rectangles) is thus perhaps a Dehn-like invariant (which has to vanish on rectangles with rational height-ratios).

More generally, given a subgroup of $SO_2$ (say finite or cyclic), can one similarly describe the obstruction for restricted scissor-equivalence allowing only (translations and) rotations in the subgroup? (The foliated case above corresponds of course to the subgroup of two elements in $SO_2$.)