Thanks! This seems to nail it indeed,

@MoisheKohan I guess you referring to "Extrinsic geometry of convex surfaces"? The only notion of rigidity I found there is infinitesimal (p. 260: a surface is said to be rigid if it admits no nontrivial bending fields).

@RobertBryant Thanks for pointing out the Hilbert and Cohn-Vossen book. In that same chapter the following statement appears: "It is likewise impossible to bend any convex surface having boundary curves, provided each boundary curve has the property that at all of its points the tangent plane is the same. An example of such a surface (with two boundary curves) is the convex part of the torus". I wasn't able to figure out whether the impossibility to bend refers here to infinitesimal deformations (which I know how to prove) or does it refer to rigidity with respect to finite deformations.

Their argument is quite straightforward and applies to general positively curved surfaces; indeed, there is a local argument which is then more or less supplements by a clopen argument to show that two immersions that agree on a segment agree everywhere. Again, this is a uniqueness clause and not an existence clause.

Thanks Robert Bryant. . The above uniqueness clause is modulo a rigid transformation. I am actually looking for a non-local answer. For example, how much flexibility is there is deforming isometrically a hemisphere? Or for which submanifolds of the sphere a smooth isometric immersion is necessarily also an embedding?

@WillieWong Yes, this is essentially the argument, but it is not so straightforward. I needed some kind og bootstrap before making this argument work. The $v_n$ being Killing fields, all norms are equivalent, so you don't even need to invoke ellpiticity,

@WillieWong If the map is everywhere orientation preserving, then no problem. But if you allow the map to change orientation then the projection on O(d) is no longer a Sobolev map

@WillieWong Just as a sketch of the proof, for every Sobolev map between manifolds of equal dimensions, div(cof df))=0 (with the appropriate interpretation)). If df is in SO(), then cof(df) = df, i.e., the map is (weakly) harmonic, hence smooth. (The complete proof involves more technical details, but that's the gist of it.)