I ask this question with some trepidation, because it may be trivial and/or of entirely recreational interest.

Erika Pannwitz proved in 1933 that every non-trivial knot contains a quadrisecant (four points which lie along a common line). Generically, a knot has finitely many quadrisecants. But what is so special about the condition of collinearity? In a dream I had a few nights ago (I apologise that this question arose in such a context), somebody asked me whether every knot contains all four vertices of an isosceles trapezoid (not in those words). I could answer the question neither in my sleep nor after I woke up. The claim sounded plausible to me (the codimension seems about right), and if it's true, then one might dream that a signed count of such trapezoids might give rise to a knot invariant along the lines of Budney-Conant-Scannell-Sinha.

Does every knot contain all four vertices of an isosceles trapezoid? More generally, is there a nice description of the subclass $\mathcal{C}$ of quadrilaterals such that every knot contains all four vertices of at least one, and generically finitely many, quadrilaterals in $\mathcal{C}$?

EDIT: Does every knot contain all four vertices of a rectangle?

I have thought about this problem a bit (trying to put it in the framework people use for dealing with colinearity problems), and it feels like it should be easy (and well-known to experts), but I'm a bit stuck.