**Setting**: Let $(M,g)$ be a smooth Riemannian manifold with Levi Civita connection $\nabla$.

**Definitions and known facts**:

1. If $\pi: E \to M$ is a vector bundle and $\sigma : [0,1] \to M$ is a smooth curve we can define the parallel transport $\mathcal{P}\_{E}^{\sigma}$ from $E\_{\sigma(0)}$ to $E\_{\sigma(1)}$.
2. If $E=TM$ it can be proven that $\mathcal{P}\_{TM}^{\sigma}$ is an isometry i.e. $$g\_{\sigma(0)}(u,v) = g\_{\sigma(1)}(\mathcal{P}\_{TM}^{\sigma}(u), \mathcal{P}\_{TM}^{\sigma}(v))$$
for every $u,v \in T\_{\sigma(0)}M$.
3. Now, fix $h, k \in \mathbb{N}$; we know that $g$ induces a bundle metric $\tilde{g}$ on $\tau^h\_k(M)$ (the space of $(h,k)$-tensor fields on $M$). With $E=\tau^h\_k(M)$we can define the parallel transport $\mathcal{P}\_{\tau^h\_k(M)}^{\sigma}$.

**My question**:

Is $\mathcal{P}\_{\tau^h\_k(M)}^{\sigma}$ an isometry? In other words, is it true that
$$\tilde{g}\_{\sigma(0)}(u,v) = \tilde{g}\_{\sigma(1)}(\mathcal{P}\_{\tau^h\_k(M)}^{\sigma}(u), \mathcal{P}\_{\tau^h\_k(M)}^{\sigma}(v))$$
for every $u,v \in (\tau^h\_k)\_{\sigma(0)}(M)$?

I think that my question is equivalent to ask if the connection $\tilde{\nabla}$ on $\tau^h\_k$ induced by $\nabla$ is compatible with $\tilde{g}$, i.e.
$$ X \tilde{g}( Y, Z) = \tilde{g} (\nabla\_X Y, Z ) + \tilde{g}( Y, \nabla\_X Z ) $$ 
for every $X \in TM$ and $Y, Z \in \tau^h\_k(M)$.