Let $\lambda = (\lambda_1, \ldots, \lambda_r)$ and $\mu = (\mu_1, \ldots, \mu_r)$ be partitions such that $\mu_j = \lambda_j +1$ for one index $j$ and $\mu_i = \lambda_i$ for all other $i$. Then there is a natural transformation $\alpha_{\mu/\lambda}: \mathbb{S}_{\lambda}(V) \otimes V \to \mathbb{S}_{\mu}(V)$, where $\mathbb{S}_{\kappa}$ denotes the $\kappa$-Schur functor; $\alpha_{\mu/\lambda}$ is unique up to scaling.
For a smooth manifold $X$, let's define a $\mu/\lambda$-differential operator to be a map $\delta$ from sections of $\mathbb{S}_{\lambda} T^{\ast} X$ to sections of $\mathbb{S}_{\mu} T^{\ast} X$ such that, for any smooth function $f$ and section $v$, we have the Leibniz rule $$\delta(f v) = \alpha_{\mu/\lambda}(df \otimes v) + f \delta(v).$$
Let's define a natural $\mu/\lambda$-differential operator to be a choice $\delta_X$ of a $\mu/\lambda$-differential operator on each manifold $X$ such that, if $\phi: X \to Y$ is a smooth map, then $\phi^{\ast} \circ \delta_Y = \delta_X \circ \phi^{\ast}$.
Are the only natural differential operators scalar multiples of the exterior derivative $d$ with $\mu= 1^{k+1}$ and $\lambda = 1^k$?
Motivation: Just curiosity. I've been trying to make $d$ sound natural this term, and one thing that I've said a lot is that it is the only thing that commutes with pullback, so I'm curious if this formalization of that this is true.