What is known about polyhedra nets that allow overlapping? It is an open problem that the net of any convex polyhedron can be unfolded onto a flat plane with no overlapping. Is anything known if we allow x faces to overlap? For example, is it known if any convex polyhedron can be unfolded with a maximum of 2 faces overlapping? What about the more general case of a polyhedron that is topologically convex (that is, its graph is isomorphic to the graph of a convex polyhedron)?
This paper provides an example of an open polyhedron without a net that is topologically convex, and the closed case appears to be taken care of by this paper (mentioned in an answer below). One solution to the topologically convex case would then to be to find a procedure to modify either polygon so that the number of overlapping sides increases without bound. I have been unable to do so without breaking the topologically convex property, but it seems a reasonable task.
 A: To the case of topologically convex polyhedra:  There is example in article by A.Tarasov: "Take a regular pyramid with base a regular non-convex dodecagon whose angles are
alternately 280 and 20 degrees, shall we say. If the vertex of the pyramid is projected onto the centre
of the base, then in the unfolding all the lateral faces must be cut o from the base or else there
is self-overlapping." Of course this pyramid is topologically convex, because its graph is the same as the graph of regular 10-sided pyramid. 
Even more, in this paper was proven, that there exists a non-convex polyhedral sphere with convex faces and having no natural unfolding(i.e. without non-overlapping net). Unfortunately, there are no pictures in this article, you can see one example of such polthedron in 
another article by Grunbaum.
A: I would like to point out a closely related question, which is described in our book that Joe Malkevitch mentioned (p.308).  We called it the Fewest Nets problem.  What is the fewest number of non-overlapping nets into which you can partition the surface of a convex polyhedron, cutting along polyhedron edges?  The answer might be 1.  But our ignorance leaves us with just fractions of F, the number of faces of the polyhedon.  The best fraction obtained to date (as far as I know) is (1/2)F.
