Is the covariant Hessian preserved by isometries? Let $(M_1,g_1,\nabla^1)$ and $(M_2, g_2, \nabla^2)$ be pseudo-Riemannian manifolds equipped with their Levi-Civita connections and $F\colon M_1 \to M_2$ be an isometry. 
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I wondered if given $f \in C^\infty(M_2)$, we have $${\rm Hess}(f\circ F)(\vec{X}, \vec{Y}) = {\rm Hess}(f)({\rm d}F(\vec{X}),{\rm d}F(\vec{Y})),$$for given $\vec{X},\vec{Y} \in \mathfrak{X}(M_1)$. I'm probably missing some subtlety on notation, but here's my attempt.
Recalling that ${\rm d}F(\vec{X})(f) = \vec{X}(f\circ F)$, etc., we have $$\begin{align} {\rm Hess}(f\circ F)(\vec{X},\vec{Y}) &= \vec{X}(\vec{Y}(f\circ F)) - (\nabla^1_{\vec{X}}\vec{Y})(f\circ F) \\ &= \vec{X}({\rm d}F(\vec{Y})(f)) - {\rm d}F(\nabla^1_{\vec{X}}\vec{Y})(f) \\ &= \vec{X}({\rm d}F(\vec{Y})(f)) - (\nabla^2_{{\rm d}F(\vec{X})}{\rm d}F(\vec{Y}))(f).\end{align}$$I don't know how to justify that $\vec{X}({\rm d}F(\vec{Y})(f)) = {\rm d}F(\vec{X})({\rm d}F(\vec{Y})(f))$. I'm not sure that this is even true, since ${\rm d}F(\vec{Y})(f) \in C^\infty(M_1)$ and ${\rm d}F(\vec{X})\in\mathfrak{X}(M_2)$. Help?
 A: Edit: skip to the end for the good solution.
I managed to solve the problem by bashing at it like a neanderthal, but I'm not happy yet. For the record:
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Take $(U,(x^1,\ldots, x^n)) \in \Sigma(M_1)$ a chart in $M_1$, and the corresponding chart $(F[U], (y^1,\ldots, y^n)) \in \Sigma(M_2)$, where $y^i \circ F = x^i$. This means that for each $j$, we have $$\frac{\partial}{\partial x^j}(f \circ F) = \frac{\partial f}{\partial y^j} \circ F,$$whence: $$\begin{align} \vec{X}(\vec{Y}(f\circ F))&= \vec{X}\left( \sum_j Y^j \frac{\partial}{\partial x^j}(f \circ F)\right) \\ &= \sum_j\vec{X}\left(Y^j\frac{\partial f}{\partial y^j}\circ F\right) \\ &= \sum_{i,j}X^i \frac{\partial}{\partial x^i}\left(Y^j \frac{\partial f}{\partial y^j}\circ F\right) \\ &= \sum_{i,j} X^i Y^j \frac{\partial^2f}{\partial y^i \partial y^j}\circ F + \sum_{i,j} X^i \frac{\partial Y^j}{\partial x^i} \frac{\partial f}{\partial y^j}\circ F. \end{align}$$On the other hand, ${\rm d}F(\vec{X}) = \sum_i (X^i\circ F^{-1}) \frac{\partial}{\partial y^i}$ gives $$\begin{align} {\rm d}F(\vec{X})({\rm d}F(\vec{Y})(f))&= {\rm d}F(\vec{X})\left(\sum_j (Y^j\circ F^{-1}) \frac{\partial}{\partial y^j}(f) \right) \\ &= \sum_j{\rm d}F(\vec{X})\left( (Y^j\circ F^{-1}) \frac{\partial f}{\partial y^j}\right)  \\ &= \sum_{i,j}(X^i\circ F^{-1}) \frac{\partial}{\partial y^i}\left((Y^j\circ F^{-1}) \frac{\partial f}{\partial y^j}\right) \\ &= \sum_{i,j} (X^i\circ F^{-1})(Y^j\circ F^{-1})\frac{\partial^2f}{\partial y^i \partial y^j} + \sum_{i,j}(X^i\circ F^{-1}) \frac{\partial}{\partial y^i}(Y^j \circ F^{-1}) \frac{\partial f}{\partial y^j} \\ &= \sum_{i,j} (X^i\circ F^{-1})(Y^j\circ F^{-1})\frac{\partial^2f}{\partial y^i \partial y^j} + \sum_{i,j}(X^i\circ F^{-1}) \left(\frac{\partial Y^j}{\partial x^i}\circ F^{-1} \right)\frac{\partial f}{\partial y^j} . \end{align}$$

Meaning $$\vec{X}(\vec{Y}(f\circ F)) = {\rm d}F(\vec{X})({\rm d}F(\vec{Y})(f)) \circ F.$$

Forgetting abuses of notation, my guess would be $${\rm Hess}(f\circ F)(\vec{X},\vec{Y}) = {\rm Hess}(f)({\rm d}F(\vec{X}),{\rm d}F(\vec{Y}))\circ F,$$which now holds.

With Aloisio's comment, here goes the non-stupid way to go at it: $$\begin{align} {\rm Hess}(f\circ F)(\vec{X},\vec{Y}) &= \vec{X}(\vec{Y}(f\circ F)) - (\nabla^1_{\vec{X}}\vec{Y})(f\circ F) \\ &= \vec{X}({\rm d}F(\vec{Y})(f)\circ F) - {\rm d}F(\nabla^1_{\vec{X}}\vec{Y})(f) \\ &= {\rm d}F(\vec{X})({\rm d}F(\vec{Y})(f))\circ F - ((\nabla^2_{{\rm d}F(\vec{X})}{\rm d}F(\vec{Y}))\circ F)(f) \\ &= {\rm Hess}(f)({\rm d}F(\vec{X}),{\rm d}F(\vec{Y}))\circ F. \end{align}$$
A: While your solution to the problem seems to work, I still think there is a point to be made here. The Hessian is fully determined by the smooth structure and Riemannian metric (as the Levi-Civita connection is). Since these structures are preserved by $F$, the equality 
$$
\mathrm{Hess}(f\circ F)=F^*\mathrm{Hess}f
$$
is a tautology.
