In the configuration space of 4 points in $\mathbb R^3$, the subspace of isosceles trapezoids has codimension three. So on a generic knot, you'd expect a 1-parameter (possibly empty) family of isosceles trapezoids whose vertices sit on the knot.
Regarding your question about inscribed rectangles, I doubt there's a way to extract an invariant of knots from this. Generically a knot has only finitely many inscribed rectangles, but 1-parameter families of inscribed rectangles can degenerate, allowing two edges to come together.
At this degeneracy you have a configuration of two tangent vectors on the knot, where the vectors are parallel and the base-points are separated by an orthogonal vector. You can check that in the configuration space of two points along the knot, such configurations are co-dimension 3. So in general, a 1-parameter family of knots can have such degeneracies, and this would allow for a 1-parameter family of inscribed rectangles to collapse, allowing for an individual inscribed rectangle to slide-off the knot via a 1-parameter family of knots.
It's possible you can find a correction-term -- when the rectangle slides off the knot, perhaps there's another corresponding thing to count. But it's not clear to me what that should be.
As a concrete example -- consider a type-2 Reidemeister move, when you have two parallel strands but before you cross them. If you were to apply a type-1 Reidemeister move to one of the strands, you would create an inscribed rectangle, one for every twist. So inscribed rectangles might be a non-diagrammatic analogue to writhe of a knot diagram.