Let $M$ be a smooth Riemannian manifold with metric $\langle\cdot,\cdot\rangle_x$, and consider a smooth function $f:M\to\mathbb{R}$. Suppose that we consider another Riemannian metric $\langle\cdot,\cdot\rangle_x'$ on $M$. Then we know that for each $x\in M$ there exists a unique positive definite linear operator $P_x:T_xM\to T_xM$ such that \begin{equation} \langle u,v\rangle_x' = \langle u,P_xv\rangle_x \qquad \forall u,v\in T_xM. \end{equation} Knowing $P_x$, one can derive the Riemannian gradient of $f$ with respect to the metric $\langle\cdot,\cdot\rangle_x'$ from the Riemannian gradient with respect to the metric $\langle\cdot,\cdot\rangle_x$ as \begin{equation} \mathrm{grad}^{\langle\cdot,\cdot\rangle_x'} f(X) = P_x^{-1}\mathrm{grad}^{\langle\cdot,\cdot\rangle_x} f(X). \end{equation} Does a similar relation hold also for the Riemannian Hessian, i.e. \begin{equation} \mathrm{Hess}^{\langle\cdot,\cdot\rangle_x'} f(X) = P_x^{-1} \circ \mathrm{Hess}^{\langle\cdot,\cdot\rangle_x} f(X)? \end{equation}
I tried by proving that $\bar{\nabla}_UV:= P_x^{-1}\nabla_UP_xV$ is the Levi-Civita connection w.r.t $\langle\cdot,\cdot\rangle_x'$, where $\nabla$ is the Levi-Civita connection w.r.t. $\langle\cdot,\cdot\rangle_x$, but I think that this is not true (I proved that $\bar{\nabla}$ is a connection, but don't see a way to prove symmetry and compatibility with the metric). Can we say something at least in the case where $M$ is an embedded submanifold of a Euclidean space $E$ and $\langle\cdot,\cdot\rangle_x= \langle\cdot,\cdot\rangle$ , $\langle\cdot,\cdot\rangle_x' = \langle\cdot,\cdot\rangle'$ are two metrics of the embedding space?