Does the curvature determine the metric?

Here's another example.

First, imagine a short cylinder $S^1\times [0,1]$ and a long cylinder $S^1\times [0,10^{10}]$, but with the same radii. Smoothly cap off both ends of the cylinders in the same way using spaces homeomorphic to discs.

The resulting manifolds are both homeomorphic to $S^2$, are not isometric (since one has a much larger diameter than the other), but there is a curvature preserving diffeomorphism between them.

To see this, just convince yourself there is a diffeomorphism $f:S^1\times[0,1]\rightarrow S^1\times [0,10^{10}]$ with the property that $f$ is an isometry when restricted to $[0,\frac{1}{4}]$ and $[\frac{3}{4},1]$. This condition allows you to extend $f$ to a curvature preserving diffeo of both compact manifolds.

Imagine a sphere with several identical bumps. More precisely, the surface given in spherical coordinates by $\rho=F(\theta,\phi)$ where $F$ is equal to one except in some places that are far one from another. To be concrete, suppose there are four bumps uniformly distributed along the equator.

You can grab one of these bumps and slide it to the North Pole, or, if you wish, arrange them at the vertices of a tetrahedron. There is a diffeomorphism that maps the initial surface onto the new one, and preserves curvature. Namely, the diffeomorphism is the isometry of corresponding bumps, extended in an arbitrary way to the rest of the sphere (where the curvature is constant).

The surfaces will not be isometric in general. For example, the diameter of the surface is greater when you have two bumps exactly opposite each other on the sphere.