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And more generally, actually:

Let $M \subset \mathbb{R}^3$ be a regular surface locally isometric to the plane and suppose a regular parametrization of $M$ is $\phi: \mathbb{R}^2 \to M$ given by $(u, v) \mapsto (x(u, v), y(u, v), z(u, v))$, that sends every point in the plane to a point of $M$. Define another map $Q: \mathbb{R}^2 \to M $ in the following manner:

$$\mathbb{R}^2 \ni p \mapsto \phi(A(p)) $$

where $A(p)$ is an arbitrary isometry of $\mathbb{R}^2$ (and also $\phi$ is chosen carefully enough that it has the same coefficients of the fundamental form as the plane). Is it true that $Q$ is an isometry of $M$? If so, can the reuslt be generalized to "images of isometries under regular parametrizations are themselves isometries"? And again, if that's true, how would I state this for riemannian manifolds in general?

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    $\begingroup$ You mean locally isometric to the plane. I mean, a cylinder is not globally isometric to the plane :) $\endgroup$ Aug 21, 2018 at 19:47
  • $\begingroup$ You're right, thanks! I've corrected it $\endgroup$ Aug 21, 2018 at 19:54
  • $\begingroup$ Here's an important hint: Is $\phi$ itself a (local) isometry? $\endgroup$ Aug 21, 2018 at 20:38
  • $\begingroup$ @TedShifrin Since they are locally isometric, yes, right (the coefficients of the first fundamental form have to be the same)? And since the composition of isometries is an isometry, we're done. Is that all? $\endgroup$ Aug 21, 2018 at 21:49
  • $\begingroup$ Well, you didn't specify that your parametrization was carefully chosen. If it's not, .... $\endgroup$ Aug 21, 2018 at 21:52

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The question has been answered in the comments.

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