Let $M$ be an oriented manifold, $GL^+(M)$ be a bundle of all positively oriented frames and $P:=GL^+(M)/SO(n)$ where the quotient is understood with respect to the natural action of $SO(n)$ on $GL^+(M)$. I would like to see the proof of the following fact:
There is no diffeomorphism invariant metric on $P$.
Here invariance mean the following: if $\varphi$ is an orientation preserving diffeomorphism it acts on $M$ in the tautological way. It also acts on $P$ via $\varphi \cdot u:=d_x\varphi \circ u$ where we view a frame $u \in P_x$ as linear isomorphism $U:\mathbb{R}^n \to T_xM$. This is well defined at the level of quotient. Also $\varphi$ acts on vectors $\xi \in T_uP$ via $\varphi \cdot \xi:=d_u\tilde{\varphi}(\xi)$ where $\tilde{\varphi}(u)=\varphi \cdot u$. Invariance of the metric $g$ should mean that $$g_u(\xi,\xi')=g_{\varphi \cdot u}(\varphi \cdot \xi,\varphi \cdot \xi')$$ for $u \in P, \xi,\xi' \in T_uP,\varphi \in Diff^+(M)$
I am not sure what you eactly mean by your question. The standard use of the bundle $P$ is that a smooth section of $P$ is equivalent to a reduction of structure group of $GL^+(M)$ to the group $SO(n)$ and hence to a Riemannian metric on $M$.
The fact that there cannot be a Riemannian metric on $P$ which is preserved by $Diff^+(M)$ simply follows since the isometry group of any Riemannian manifold is a finite dimensional Lie group, whereas $Diff^+(M)$ certainly is infinite dimensional. (And there are many arguments of similar flavor, say that an isometry of a connected manifold is determined globally by its one-jet in a point, while you can change deiffeomorphisms globally without changing them on some open subset.)