View $h_i$ as conjugate linear bundle isomorphisms $h_i:E\to E^*$. Then $\phi=h_1^{-1}\circ h_2$ satisfies $h_2(x,y)=h_1(\phi.x,y)$.

View $h_i$ as conjugate linear bundle isomorphisms $h_i:E\to E^*$. Then $\phi=h_1^{-1}\circ h_2$ satisfies $h_2(x,y)=h_1(\phi.x,y)$. Since $h_2$ is positive hermitian, the endomorphism $\phi$ is positive hermitian with respect to $h_1$, thus $\sqrt{\phi}$ exists with respect to $h_1$ and $h_2(x,y)=h_1(\sqrt{\phi}.x,\sqrt{\phi}.y)$.

View $h_i$ as conjugate linear bundle isomorphisms $h_i:E\to E^*$. Then $\Phi=h_2\circ h_1^{-1}$ does it.

View $h_i$ as conjugate linear bundle isomorphisms $h_i:E\to E^*$. Then $\phi=h_1^{-1}\circ h_2$ satisfies $h_2(x,y)=h_1(\phi.x,y)$.