This is a cross-post.
It seems to be folklore knowledge that all the (source) symmetries of the $d$-Dirichlet energy are conformal maps.
Specifically, I have found this nice proof for the following claim:
Proposition: Let $M,N$ be oriented $n$-dimensional Riemannian manifolds, let $f:M \to N$ be a (local) diffeomorphism. Suppose that for any $h \in C^{\infty}(N)$, $$ E_N^p(h)=E_M^p(h \circ f)$$ where $E^p$ is the $p$-energy: $E_N^p(h)=\int_N \|dh\|^p \operatorname{Vol}_N$. (we assume $p\ge 1$).
Then,
If $p \neq n$, $\,f$ is an isometry. If $p = n, \,f$ is conformal.
Question: Are there any other proofs around?
While the proof I mentioned is nice, there are some "magical choices" involved, and I wonder if there is a simpler proof (with less computations).
The challenge is to pass "from global to local"; Here is a naive approach:
Suppose $f:M \to N$ is an orientation-preserving diffeomorphism:
We assume $$\int_M \|dh \circ df\|^p \operatorname{Vol}_M=\int_N \|dh\|^p \operatorname{Vol}_N=\int_M (\|dh \|^p \circ f ) \det df \operatorname{Vol}_M. \tag{1}$$
for every $\, h \in C^{\infty}(N).$
Since $h$ can be chosen arbitrarily we hope to deduce equality of the integrands, i.e
$$ \|dh \circ df\|^p = (\|dh \|^p \circ f ) \det df \, \, \text{for every} \, h \in C^{\infty}(N). \tag{2}$$
Is there a way to deduce this? (from here the "pointwise question" is easy).
Selecting $h$ to be very "localized" and using a Lebesgue differentiation-type theorem seems a reasonable idea, but I could not implement this successfully. We can try selecting $h$ which are quickly decaying, and are constant outside a small set. But then at the "transition phase" their differential will have a non-negligible integral.
Note that if we assume that $f$ restricted to arbitrarily small open subsets preserve the energy, then the proof is trivial. This is essentially an equivalent phrasing of the question: Why are global symmetries of smooth functionals are also local symmetries?
