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 $\phi^k\wedge \psi^{n-k} = \Lambda^{n-k}g^{-1}(\star\phi^k,\psi^{n-k}).vol(g)$ where $\Lambda^{n-k}g^{-1}$ is the induced inner product on $\Lambda^{n-k}T^*(G/H)$. So $\star$ is $G$-equivariant.

See 25.11 and 28.2, 28.3 of here.

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 $\phi^k\wedge \psi^{n-k} = (\Lambda^{n-k}g^{-1})(\star\phi^k,\psi^{n-k}).vol(g)$ where $\Lambda^{n-k}g^{-1}$ is the induced inner product on $\Lambda^{n-k}T^*(G/H)$. 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 $s:G\to \lbrace-1,1\rbrace$ 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.