Suppose that we have a convex cap, i.e., a convex surface in $R^3$ homeomorphic to a disk whose boundary lies in a plane. Reflect the cap through the plane of its boundary and glue it back to the original cap along the common boundary curve. If the closed surface we obtain is convex, then it is rigid by a theorem of Pogorelov, i.e., any convex surface which is isometric to it is congruent to it.
My question is: do we still have rigidity when the doubled surface is not convex? I do not know the answer even in the case where each cap is spherical. Here we want to look at isometric surfaces which are smooth except along the curve where the two pieces meet, so that we do not get trivial examples by reflecting a portion of the surface.
In general very little is known about rigidity of non-convex surfaces, but this situation is so simple that perhaps someone might have an idea about it. It seems to me that this is the simplest non-convex case.
Addendum: There is a theorem of Alexandrov and Sen'kin, see p. 181 in Pogorelov, which seems to be relevant to the above problem. This theorem states that if a pair of isometric convex surfaces lie in the upper half-space, are star-shaped and concave with respect to the origin, and their corresponding boundary points are equidistant from the origin, then they are congruent.