TITLE: All compact surfaces $S\subseteq \mathbb{R}^3$ are rigid?
QUESTION [20 upvotes]: Recently I've come across a lecture in differential geometry by Fernando Codá (in portuguese!) in which he stated that the following problem is (at least, up to 2014) open:

Given $S\subseteq\mathbb{R}^3$ a smooth (i.e. $C^\infty$) compact boundaryless connected surface (therefore orientable), let $f:S\times[0,1]\to \mathbb{R}^3$ be a differentiable map with $S_t:=f(S\times\{t\})$ being a surface and $f_t:S\to S_t$ an isometry (where $f_t(p):=f(p,t)$), where $f_0\equiv \mathrm{Id}_S$. Then there are rigid motions $A_t:\mathbb{R}^3\to\mathbb{R}^3$ with $f_t(S) = A_t(S)$.

More generally, in the cases of the sphere $S^2$ (Liebmann's Theorem) and "ovaloids", i.e. surfaces with Gaussian curvature $K$ strictly positive (Cohn-Vossen's rigidity theorem), it is known that isometric embeddings of it in $\mathbb{R}^3$ are exactly the restrictions of the rigid motions from $\mathbb{R}^3$. Not all compact connected surfaces $S\subseteq\mathbb{R}^3$ satisfies this, as pointed out in the book Curves and Surfaces, by S. Montiel and A. Ros:

Anyway, it is possible to show that this isometry in particular cannot be obtained by a continuous family of isometries from the identity map to the latter, so it's not a counterexample to the conjecture.
Although this seems to me to be a very studied problem, and I imagine that there must be several approachs to it, I just couldn't find enough information about it on the internet (the Montiel & Ros book was recommended by Codá later in the same lecture linked above), for "rigidity" seems to be a very comprehensive term. So my actual question is:

Actual question: Where can I find more information about this problem and the current approaches going on?

REPLY [5 votes]: Not at all an answer: I once had a conversation about this question with a friend at IHES, and Gromov was hanging around, overheard us, and said: oh, that's a stupid problem!
We said (in one voice): why do you say that?
To which Gromov responded: because the question has been around for a century, and no interesting mathematics has come out of it.
Make of this what you will.