Q:

Is parallel transport on the space of tensors an isometry?

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

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 $E_{\sigma(1)}$.
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 $u,v \in T_{\sigma(0)}M$.
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 $Y, Z \in \tau^h_k(M)$.

A:

If you assume that $\nabla$ on your vector bundle $E$ is metric compatible to begin with, then you should think of $\nabla$ as infinitesimal parallel transport. Actually I think you already do, because you brought up metric compatibility yourself.
If $Y_0$ and $Z_0$ are vectors at $\sigma(0)$, with parallel-transported vector fields $Y_t$ and $Z_t$ along $\sigma(t)$ (so e.g. $\nabla_{\sigma'(t)}Y_t = 0$), then metric compatibility implies (maybe with bad notation)$$\frac{\text{d}}{\text{d}t}g(Y_t,Z_t) = \sigma'(t) ~ g(Y_t,Z_t) = g(\nabla_{\sigma'(t)}Y_t,Z_t) + g(Y_t,\nabla_{\sigma'(t)}Z_t) = 0$$
So the quantity $g(Y,Z)$ is constant as you go along $\sigma(t)$.
Conversely, if the parallel transport of some connection is always an isometry (i.e. $g(Y,Z) = g(\mathcal P_t Y,\mathcal P_t Z)$ for all $t$), then I am pretty sure that you can say the connection is metric compatible (using some equation that roughly looks like $\nabla_X \approx \frac{\text{d}}{\text{d}t}\mathcal P_t$, for $\mathcal P_t$ along a path with velocity $X$).
If you're asking specifically if $\nabla$ is compatible with the induced metric on the tensor bundles, then yes. The induced $\tilde g$ looks like copies of $g$ and $g^{-1}$ tensored together. Since metric compatibility is the statement that $\nabla g = 0$, we also have $\nabla \tilde g=0$ by using the "Liebniz rule" definition of the induced connection $\nabla$ on tensors.
edit: As an example, on simple $(0,2)$-tensors, $\tilde g = g \otimes g$ uses ordinary multiplication, and acts like $$\tilde g(a \otimes b , c \otimes d) = g(a,c)g(b,d).$$
On 1-forms (a.k.a. $(1,0)$-tensors), $\tilde g = g^{-1}$ to preserve the natural pairing with vectors. So on say $(1,1)$-tensors, $\tilde g = g\otimes g^{-1}$, using multiplication again, and of course you can jump to the $(p,q)$ case: $\tilde g = g\otimes \dotsm\otimes g\otimes g^{-1} \otimes \dotsm \otimes g^{-1}$. 
I guess since $g$ is a $(2,0)$-tensor field, $\tilde g_{(p,q)}$ is a $(2q,2p)$-tensor field.
Now $\tilde g$ is parallel - take the $(2,0)$-case again: $$\nabla \tilde g = \nabla g \otimes g + g \otimes \nabla g = 0$$ from which you can also figure out the $(p,q)$ case.
I hope that answers it.

