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Let $(M,g), (\overline{M},\overline{g})$ be smooth Riemannian manifolds that are isometric, i.e. there is a smooth function $f: M \rightarrow \overline{M}$ such that d$f(g)=\overline{g}$. By Nash's embedding theorem, there are smooth isometric embeddings $e: M \rightarrow \mathbb{R}^n$, $\overline{e}: \overline{M} \rightarrow \mathbb{R}^n$ for some $n$. Does there always exist an element $X \in E(n)$, i.e. a combination of rotations, reflections and translations of $\mathbb{R}^n$, such that $X\big(e(M)\big)=\overline{e}(\overline{M})$?

Here, $E(n)$ is the group of isometries of $\mathbb{R}^n$. It is generated by rotations, reflections and translations.

Addendum: I guess we should require the mean curvatures at the corresponding points be equal.

Has
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    Well, for instance, we can isometrically embed an open interval in the plane in two different ways – a straight line and an arc of a circle... – Zhen Lin Sep 19 '15 at 13:03
  • Oops. Why are some spaces more rigid than others? For example, any surfaces in $\mathbb{R}^3$ isometric to the upper half sphere should be related through $E(3)$ right? – Has Sep 19 '15 at 13:55
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    @Has: Take a look at "Pogorelov's uniqueness theorem", http://math.stackexchange.com/questions/460823/isometries-of-the-sphere. This theorem shows that sphere with a metric of positive curvature admits unique isometric embedding into $R^3$. As for your question "Why are some spaces more rigid than others", the brief answer is "curvature": Positive curvature tends to force rigidity; zero curvature (like in the case of an arc) tends to force flexibility. – Moishe Kohan Sep 25 '15 at 00:40

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