Question: the gradient of a function $f:E\rightarrow\mathbb{R}$, with $E$ being a vector space equipped with an inner product $\langle\cdot,\cdot\rangle$, depends on the choice of the inner product itself. What is the influence of this choice on the Hessian matrix of $f$? In particular, if $E=\mathbb{R}^n$, what is the link between the "regular" Hessian matrix and the one computed with an arbitrary inner product?
Context: it seems clear to me how the gradient depends on the choice of inner product: the derivative of $f$ computed at $x$ is a linear function $Df(x):E\rightarrow\mathbb{R}$, and so it is an element of the dual space $E^*$. We then have an isomorphism $\Phi:E\rightarrow E^*$ induced by the inner product as $\Phi(x)(y) = \langle x,y\rangle$, which allows us to identify the derivative with the gradient by $\nabla f(x) = \Phi^{-1}(Df(x))$.
I'd like to apply the same reasoning to the second derivative, but I realized I have no clue about where the inner product comes into play. Following the same path, I expect that there is some connection between the second derivative $D^2f(x):E\times E\rightarrow\mathbb{R}$ (which shouldn't depend on the inner product) and the derivative of the gradient $D\nabla f(x): E\rightarrow E$ (sorry for the notation!).
Edit based on the helpful comment by @NickAlger: what I call $D\nabla f(x)$ can be written as $\Phi^{-1}\circ H_x: E\rightarrow E$, where $H_x$ denotes the Hessian computed at $x$, i.e., the linear operator mapping $E$ to its dual through $(H_xu)(v) = D^2f(x)(u,v)$.
I'm interested in the matrix representation of $(\Phi^{-1}\circ H_x)$ for different choices of the inner product. Since $\Phi$ is always a linear isomorphism, I think this all boils down to taking different bases for domain and image, but I have a hard time formalizing this idea.
Any help would be appreciated. Thanks!