Perhaps this is a naive question (and certainly an idle one):
If $\phi:\mathbb{R}^3\to \mathbb{R}^3$ is a smooth diffeomorphism with the property that for any compact surface $\Sigma\subset \mathbb{R}^3$ one has
$$Area(\phi(\Sigma))=Area(\Sigma)$$
is $\phi$ necessarily an isometry? Here we are using the euclidean metric.
If the map is distance preserving then it is an isometry by a theorem of Myers and Steenrod (see here for instance). If the map is only volume preserving then it could be affine (or worse) so the result couldn't hold.
I had thought perhaps to approximate geodesics by thin tubes but couldn't see how any such argument wouldn't also hold for volume preserving maps.
