Number of distinct nets of dual polyhedra There are 11 non-congruent nets of a cube as well as 11 distinct nets of an octahedron. Both a dodecahedron and an icosahedron have 43380 distinct nets.
Is it true that any pair of dual convex polyhedra always have the same number of distinct nets (let's forget about self-intersections for simplicity)? I believe, there should be some simple proof.
 A: I am not sure how to define a net in general, but if we disregard self-intersection, a net should correspond to a choice of edges that are cut, so that the faces are still connected, but there are no circles (which would prohibit the flattening of the net).
So if I regard the polyhedron as a graph, a net is a choice of edges=dual edges to remove, so that the dual graph becomes a tree. So I think that you are looking at spanning trees of graphs and their duals. 
And the complement of the edges of a spanning trees viewed as dual edges form a spanning tree of the dual graph, so their number is equal.
(In particular, if you regard the corresponding edges of the cube and the octahedron, each net of the cube corresponds to a net of the octahedron where one has cut open exactly those edges that have not been cut in the cube.)
A: At first, I could not understand the relevance of Joseph Malkevitch's answer. After investigating, however, I think I know what he was getting at.
Phira's answer is correct: Nets correspond to spanning trees on the vertex-edge graph of the polyhedron $P$, and each spanning tree $T$ is paired with a spanning tree $T^*$ in the dual graph. The nodes of the dual graph are the faces of $P$, and $T^*$ has edges crossing each edge of $P$ which is NOT in $T$.
This sets up a one-to-one correspondence between nets for $P$ and nets for $P^*$. Moreover, since $P$ and $P^*$ have the same symmetry group, the same symmetry operations which create a congruency class of nets for $P$ also create a congruency class of nets for $P^*$. So, usually, there are the same number of non-congruent nets for a polyhedron and its dual.
However, it's possible to have two nets which are congruent "by coincidence", and not due to any symmetries of the source polyhedron. In this case, the corresponding nets for the dual polyhedron may not be congruent.
An example occurs with the triangular prism and its dual, the triangular bipyramid. The prism has 9 non-congruent nets; the bipyramid has only 8.
Here are all the nets, paired up:

You can see that nets 5 and 9 for the triangular bipyramid are congruent, even though they correspond to non-congruent nets for the prism.
A: Look at the case of two regular tetrahedra pasted together along an equilateral triangle face (bi-pyramid) vs. a triangular prism, with squares and equilateral triangles, and do the cuts for each of these convex polyhedra along edges to get their "nets."
