Reference request: Naturality of Levi-Civita connection in $T^kTM$

Question

If $(M,g)$ is a Riemannian manifold, the Levi-Civita connection is the unique connection $\nabla$ on $TM$ for which the following conditions hold:

• Compatibility: $\nabla_X \langle Y, Z \rangle = \langle \nabla_X Y, Z \rangle + \langle Y, \nabla_X Z \rangle$
• Symmetry: $\nabla_X Y - \nabla_Y X = [X,Y]$

The expression $\nabla_X \langle Y, Z\rangle$ makes sense because $\nabla$ induces a connection in every tensor bundle $T^{(k,\ell)}TM$—including $T^{(0,0)}TM \cong C^\infty(M)$—for which the following properties hold:

1. $\nabla_X f = Xf$ for every $f \in C^\infty(M)$;
2. $\nabla_X (F \otimes G) = (\nabla_X F) \otimes G + F\otimes(\nabla_X G)$;
3. $\nabla_X(\mathrm{tr} F) = \mathrm{tr}(\nabla_X F)$.

The Levi-Civita connection is natural by virtue of the following result:

If $(M,g)$ and $(\tilde M, \tilde g)$ are Riemannian manifolds with Levi-Civita connections $\nabla$ and $\tilde\nabla$ respectively, and $\phi : M \to \tilde M$ is an isometry, then $\phi^*\tilde\nabla = \nabla$, where $\phi^*\tilde\nabla$ is the pullback connection defined by $$ \left(\phi^*\tilde\nabla\right)_X Y = \left(\phi^{-1}\right)_*\left(\tilde\nabla_{\phi_*X}\left(\phi_*Y\right)\right) $$

My question: Is there a similar naturality result for the Levi-Civita connection in the covariant $k$-tensor bundles?

The intuitive way to define the pullback connection in $T^kTM$ under a diffeomorphism $\phi : M \to \tilde M$ seems to be $$ (\phi^*\tilde\nabla)_X F = \phi^*\left(\tilde\nabla_{\phi_*X}\left(\phi^{-1}\right)^*F \right) $$ It's straightforward to confirm that this is in fact a connection on $T^kTM$. So my question is really, if $\phi$ is an isometry, is $\phi^*\tilde\nabla = \nabla$?

I'm not sure how one would prove this. The naturality argument for the Levi-Civita connection in $TM$ is based on showing $\phi^*\tilde\nabla$ on $TM$ is both symmetric and compatible with the metric $g$, and invoking uniqueness of the Levi-Civita connection. There's a compatibility result for arbitrary (mixed) tensor bundles analogous to the compatibility property in $TM$: $$ \nabla_X\langle F, G \rangle = \langle \nabla_X F, G\rangle + \langle F, \nabla_X G\rangle $$ However there doesn't seem to be as obvious an analogous symmetry result.

Furthermore, one would have to show the Levi-Civita connection $\nabla$ in $T^kTM$ is the unique connection compatible with $g$ in the tensor field sense. To prove this, if one were to use uniqueness in $TM$, it seems to me one would have to recover $\nabla$ on $TM$ from $\nabla$ on $T^kTM$, which doesn't seem easy either.

Is there a reference for a result like this someone could point me towards?

EDIT: Yes, if $\phi : M \to \tilde M$ is an isometry and $\nabla$ and $\tilde\nabla$ are the connections in $T^kTM$ and $T^kT\tilde M$ induced by the Levi-Civita connections $\nabla$ and $\tilde\nabla$ on $TM$ and $T\tilde M$, then using the pullback definition for $\tilde\nabla$ on $T^kTM$ and $T^kT\tilde M$ as defined above, we do have have $$\phi^*\tilde\nabla = \nabla.$$ This follows from the following expansion of $\nabla_X F$ for a covariant $k$-tensor field $F$ and vector fields $X$ and $Y_1, \ldots,Y_k$: $$ (\nabla_X F)(Y_1, \ldots, Y_k) = X(F(Y_1, \ldots, Y_k)) - \sum_{i=1}^k F\left(Y_1, \ldots, \nabla_X Y_i, \ldots, Y_k\right) $$