For a homogeneous space $G/H$, endowed with a $H$equivariant metric $g$, let $\ast$ be the corresponding Hodge star map. It seems that $\ast$ must also be $\ast$equivariant, but I can't see how one would prove it. I badly recall that the Hodge star map can constructed as contraction with the uniquely determined highest form $\Omega$, and so, I guess that since contracting must be equivariant, the Hodge must be. But please can someone direct me to a good source for all this.

A $G$invariant metric $g$ on $G/H$ is uniquely determined by its $H$invariant value $g_o$ at $T_o(G/H)$ for the base point $o\in G/H$. The Riemannian volume form $vol(g)$ is $G$invariant, and $\star$ is given by $\phi^k\wedge \psi^{nk} = (\Lambda^{nk}g^{1})(\star\phi^k,\psi^{nk}).vol(g)$ where $\Lambda^{nk}g^{1}$ is the induced inner product on $\Lambda^{nk}T^*(G/H)$. So $\star$ is $G$equivariant. See 25.11 and 28.2, 28.3 of here. Edit: I was a little too fast. As Robert remarked, this is okay if $G$ also preserves the orientation. If $G/H$ is not orientable then one may go to the orientable double cover of $G/H$ where the the Hodge map exists and exchanges "formes pairs" (in the sense of the De Rham) which are invariant under the covering map, and "formes impaired", the eignespace of eigenvalue 1 under the covering map. If $G/H$ is orientable but $G$ does not respect the orientation, then $G$ respects the volume density, and there is a homomorphism $s:G\to \lbrace1,1\rbrace$ such that $g^*vol(g)= s(g).vol(g)$ and one can use that. @Mihail: You are right. If you view $\Lambda^{k}g^{1}:\Lambda^{k}T^\star M\to (\Lambda^{k}T^\star M)^\star =\Lambda^kTM$ then $\star \phi^k = i(\Lambda^{k}g^{1}(\phi^k)) vol(g)$  in the orientable case. See also this book by Friedrich and Agricola. 

