$\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$
Let $\M,\N$ be smooth oriented $d$-dimensional Riemannian manifolds. Let $f:\M \to \N$ be smooth, and let $\delta=d^*$ be the adjoint of the exterior derivative.
Supopse that:
$f^*\delta \omega=\delta f^*\omega$ for every differential form $\omega$ on $\N$ of every degree.
$df$ is invertible everywhere on $\M$.
Is it true that $f$ is an isometry?
In dimesnion $d=1$ I was able to prove this with an easy computation. For $d \ge 2$ the calculation seems to get more involved.
Note that up to signs $\delta=\star d \star$. Can we show any such $f$ commutes with the Hodge dual? If this is the case, then $f$ must be an isometry.
Comment: I am not sure the requirement that $df$ is invertible is necessary. Indeed, if $\omega$ is a $1$-form on $\N$, then $\delta \omega \circ f=f^*\delta \omega=\delta f^*\omega$, so the LHS does not involve an action by the differential of $f$, while the RHS does involve such an action. In particular, the constant map does not satisfy the equation. (For dimension $1$, the invertibility assumption is indeed not needed).
Edit:
Here is a partial answer. I still hope there is a more slick argument.
First, the condition $f^*\delta \omega=\delta f^*\omega$ for one forms implies $f$ is a harmonic morphism, i.e. it preserves harmonicity of functions under precomposition. It is known that such maps, between equidimensional manifolds, must be weakly conformal, that is at every point $x \in \M$, the differential $df_x$ is either a linear conformal map or zero. So, if we assume $df \neq 0$, we are done; Our map is conformal, and now it is not hard to see it must be an isometry.
Can we relax the requirement $df \neq 0$?
