Orientation is equivalent to an SO(n)-reduction

The orthonormal frames form a principal $\mathrm{O}(n)$-subbundle of $\mathrm{Fr}(\xi )$: Gram–Schmidt in a bundle chart gives local orthonormal frames, and any two differ by a unique orthogonal matrix. Given an orientation, let $Q$ consist of its positive orthonormal frames. In oriented orthonormal charts, $Q\cong U\times \mathrm{SO}(n)$, and every orthonormal frame is a positive one followed by an element of $\mathrm{O}(n)$. Thus $Q$ is an $\mathrm{SO}(n)$-reduction in the sense of [F1].

Conversely, let $Q$ be an $\mathrm{SO}(n)$-reduction. At $x$, transport the standard orientation of ${\mathbb{R}}^n$ through any $q\in Q_x$. Replacing $q$ by $qh$ with $h\in \mathrm{SO}(n)$ does not change the orientation. A local section of $Q$ makes the choice continuous, so it defines an orientation of $\xi$.

Starting from an orientation, step 2.1 applied to its positive orthonormal frames returns that orientation. Starting from $Q$, the positive orthonormal frames for the resulting orientation are exactly $Q$, since each $\mathrm{O}(n)$-fiber has precisely the determinant-positive coset. Hence the constructions are inverse.

An orientation-preserving isometry of metric bundles carries positive orthonormal frames to positive orthonormal frames and hence carries the associated reduction to the associated reduction. When $n=0$, both groups and the frame fiber are singletons, and the same conclusion holds.