Why is a diffeomorphism an isometry if and only if it commutes with the Laplacian?

Here's one way to make it precise.

Theorem. Suppose $(M_1,g_1)$ and $(M_2,g_2)$ are Riemannian manifolds, $\Delta_1,\Delta_2$ are their respective Laplace operators, and $\phi\colon M_1\to M_2$ is a diffeomorphism. Then $\phi$ is an isometry if and only if for every $f\in C^\infty(M_2)$ we have $\Delta_1(f\circ\phi) = (\Delta_2 f)\circ \phi$.

(Of course, if you're only interested in diffeomorphisms from a Riemannian manifold to itself, you can take $M_1=M_2=M$ and $g_1=g_2=g$.)

I don't have time to write down a complete proof, but the idea is that the principal symbol of a differential operator is a diffeomorphism-invariant quadratic function on the cotangent bundle, and the principal symbol of the Laplace operator is given by the squared norm function on covectors. By the polarization identity, the squared norm function determines the metric on covectors (the "dual metric"), and by the musical isomorphism between $TM$ and $T^*M$, this in turn determines the metric.