# deformation of Hodge star operator and harmonic forms

Suppose $$(M,g)$$ is a compact Riemannian manifold, and $$*_g$$ is the Hodge star operator defined on the de Rham algebra $$\Omega^*(M)$$ with respect to the metric $$g$$. Let $$\phi:M\to M$$ be a diffeomorphism. Then, can we find another metric $$h$$ (which should be related to $$\phi$$) so that $$(\phi^{-1})^* \circ *_g\circ \phi^*$$ agree with the new Hodge star operator $$*_h$$?

EDIT: I guess this should be the Hodge star operator for the pullback metric. On the other hand, we know $$\phi$$ induces a map $$\phi^*:H^*(M)\to H^*(M)$$ on de Rham cohomology groups, and meanwhile we also know for any metric $$h$$ there is the so-called Hodge isomorphism $$\mathcal H^*_h(M) \to H^*(M)$$ where $$\mathcal H^*_h(M)$$ denotes the $$h$$-harmonic forms. Now another interesting question is that can we find a metric $$h$$ so that we can induce a map $$\phi^*: \mathcal H_g(M) \to \mathcal H_h(M)$$?

$$\alpha\wedge{*}\beta=<\alpha,\beta>\Omega_{g}$$, this is the definition of hodge star.
Pullback by $$\varphi$$, we get:
$$LHS=\varphi^{*}(\alpha\wedge{*_{g}}\beta)=\varphi^{*}\alpha\wedge\varphi^{*}{*_{g}}\beta$$
$$RHS=\varphi^{*}(<\alpha,\beta>\Omega_{g})=\varphi^{*}<\alpha,\beta>\Omega_{\varphi^{*}g})=<\varphi^{*}\alpha,\varphi^{*}\beta>\Omega_{\varphi^{*}g}=\varphi^{*}\alpha\wedge*_{\varphi^{*}g}\varphi^{*}\beta$$.
Since $$\alpha,\beta$$ are arbitrarily given, compare both sides we get the conclusion.