Let $(M,g)$ be some smooth, Riemannian manifold. Let $d$ be the exterior derivative and $\delta$ the codifferential on forms. For a smooth vector field $X$, let $L_X$ be the Lie derivative associated to $X$. We know from Cartan formula that $L_X = d \iota_X + \iota_X d$ where $\iota_X$ is the interior derivative associated to the vector field $X$. So it is well-known that $L_X$ and $d$ commute: for any arbitrary form $\omega$, we have that $L_Xd\omega = dL_X\omega$.

This is, of course, not true for codifferentials. In general $[L_X,\delta]\neq 0$. For certain cases the answer is well known: if $X$ is a Killing vector ($L_Xg = 0$) then since it leaves the metric structure in variant, it commutes with the Hodge star operator, and so $L_X$ commutes with $\delta$. Another useful case is when $X$ is conformally Killing with constant conformal factor ($L_X g = k g$ with $dk = 0$). In this case conformality implies that the commutator $[L_X,*] = k^\alpha *$ where $\alpha$ is some numerical power depending on the rank of the form it is acting on (I think... correct me if I am wrong), so we have that $[L_X,\delta] \propto \delta$.

So my question is: "Is there a general nice formula for the commutator $[L_X,\delta]$?" If it is written down somewhere, a reference will be helpful. (In the Riemannian case, by working with suitable symmetrisations of metric connection one can get a fairly ugly answer by doing something like $\delta \omega \propto \mathop{tr}_{g^{-1}} \nabla\omega$ and use that the commutators $[L_X, g^{-1}]$ and $[L_X,\nabla]$ are fairly well known [the latter giving a second-order deformation tensor measuring affine-Killingness]. But this formula is the same for the divergence of arbitrary covariant tensors. I am wondering if there is a better formula for forms in particular.)

A simple explanation of why what I am asking is idiotic would also be welcome.