Exterior derivative as only (up to multiple) natural operator $\Lambda ^kT^\ast \rightsquigarrow \Lambda ^{k+1}T^\ast$

Question

In Kolar, Michor, & Slovak's book Natural Operations in Differential Geometry, it is proved the exterior derivative is universal in the following sense.

Proposition 25.4. For $k>0$ all natural operators $\Lambda ^kT^\ast \rightsquigarrow \Lambda ^{k+1}T^\ast$ are constant multiples of the exterior derivative.

For relevant definition see e.g this question. However, being a natural operator turns out to be an equivariance condition, which does not seem as obviously desirable as commuting with diffeomorphisms. (Maybe this is equivalent somehow? I haven't had a chance to decode what jet-group equivariance really means.)

On the other hand, a result of Palais says the the exterior derivative is the only (apriori assumed) linear map which commutes with diffeomorphisms.

So yeah, I guess I'm just trying to understand the relationship between being a natural operator in the sense of the linked question, to pleasantly commuting with all diffeomorphisms.

Answer 1

Naturality here means $f^*d\phi = d f^*\phi$ for any smooth map, diffeomorphism or not. This is formally stronger than requiring this just for diffeos. Eventually the two conditions are equivalent, after checking the results.

Edit:

Answer to Arrows comment.

If you require commuting only with pullbacks with diffeomorphisms, then even nonlinear operators of this kind $\Lambda^k T^*\to \Lambda^{k+1}T^*$ are multiples of the exterior derivative (see 25.4). But this commutes even with general pullbacks. For $k=0$, all natural operators are of the form $g\mapsto \phi(g).dg$ for some arbitrary smooth function $\phi:\mathbb R\to \mathbb R$ (see 25.5).