# Hodge star is conformally invariant on $\Lambda^{n/2}(V)$, for $n$ even

I am studying the Hodge star operator for the first time. I am trying to prove that for $$n$$ even, then for any $$\omega \in \Lambda^{n/2}(V)$$ $$\star_g \omega= \star_{\tilde{g}} \omega$$, where $$g$$ and $$\tilde{g}$$ are conformal ($$g=\lambda \tilde{g}$$ for $$\lambda >0$$), it seems obvious but I don't really know where to start.
My definition of Hodge star is: for any $$\omega, \mu \in \Lambda^k(V)$$, $$e_1, \dots , e_n$$ a positively oriented orthonormal basis of $$V$$ wrt the metric $$g$$ $$\omega \wedge \star \mu = g(\omega,\mu) e_1 \wedge \dots \wedge e_n$$ Thanks!

It suffices to check just on an orthonormal basis. If $$e_1,\dots,e_n$$ is an oriented $$\tilde g$$-orthonormal basis, then $$\bar e_1=e_1/\sqrt{\lambda}, \dots, \bar e_n=e_n/\sqrt{\lambda}$$ will be an oriented $$g$$-orthonormal basis. With $$I$$ an ordered multiindex of length $$n/2$$, let $$I'$$ denote a complementary ordered multiindex so that $$\tilde\star e_I = e_{I'}$$. It follows, then, that $$\star\bar e_I = \bar e_{I'}$$; in other words, $$\dfrac1{(\sqrt\lambda)^{n/2}}{\star} e_I = \dfrac1{(\sqrt\lambda)^{n/2}}e_{I'}$$, which means that $$\star e_I = e_{I'}$$, as before.