Does the $O(n)$ bundle of a manifold depend on the metric? Let $g_1$ and $g_2$ be two Riemannian metrics on a manifold $M$. These induce two $O(n)$ bundles on $M$, whose fibers over each point $x\in M$ are the groups of orthogonal transformations of $T_x M$ with respect to $g_1$ or $g_2$, respectively. Are these two $O(n)$ bundles necessarily isomorphic?
 A: Yes and yes: If you denote by $O^{g} M$ and $O^{h} M$ the bundles of orthonormal frames of $M$ with respect to two different metrics (relabeled as $g$ and $h$ here instead of $g_1$ and $g_2$), these bundles are isomorphic as principal $O(n)$-bundles. Nevertheless, these bundles depend on the metrics and the isomorphism between them is not canonical. 
An isomorphism $O^g M \to O^h M$ can be constructed as follows: First, find an isomorphism field $\alpha_{g, h} \in \Gamma(Iso(M))$ satisfying
$$ \forall x \in M: \forall v, w \in T_x M: g(\alpha_{g,h}(v),w) = h(v, w) $$
Since $g$ is positive definit, such a field is unique, if it exists and one can thus prove existence by constructing it locally. If you use the Gram-Schmidt process to $g$-orthonormalize a local $h$-ONF on some open neighbourhood $U$, you get a map $\xi_{g,h}:O^h U \to O^g U$ and if you denots its $g$-adjoint by $\xi_{g,h}^{*_g}$, the map $\alpha_{g,h} = \xi_{g,h}^{*_g} \circ \xi_{g,h}$ will do. 
Next, you define another isomorphism field by $b_{g,h} := a_{g,h}^{-1/2} \in Iso(M)$. Here $\_^{1/2}$ denotes taking the square root of a positive isomorphism. Finally, the map $c_{g,h}:O^g M \to O^h M$, $b=(b_1, \ldots, b_n) \mapsto (b_{g,h}b_1, \ldots, b_{g,h}b_n)$, gives the identification $O^g M \to O^h M$.
This identification is famous, because it is needed in the identification of the spinor bundles $\Sigma^g M$ and $\Sigma^h M$ of $M$ (in case $M$ admits a spin structure). This is discussed further in 'Generic Metrics and Connections on Spin- and Spinc-Manifolds' by Stephan Maier.
