= Solution
An <orthogonal structure on a real vector bundle> $E$ of rank $r$ is a choice of local frames whose transition matrices lie in the <orthogonal group> $O(r)$. In such frames declare
$$
h\left(\sum_i a_ie_i,\sum_i b_ie_i\right)=\sum_i a_ib_i.
$$
Orthogonal transitions preserve this expression, so it defines a smooth positive-definite <fiber metric>. Conversely, a <fiber metric> turns any local frame into a smooth orthonormal frame by the <Gram-Schmidt process>; the resulting transition matrices are orthogonal. The two constructions are inverse, giving the required equivalence.
An <orthogonal local trivialization> $E|_U\cong U\times\mathbb R^r$ identifies each fiber isometrically with standard Euclidean space. Equivalently, its coordinate frame satisfies $h(e_i,e_j)=\delta_{ij}$. This is local and does not assert the existence of a global frame.
Every real <vector bundle> over a <smooth manifold> admits a <fiber metric>. Choose a trivializing open cover and local Euclidean metrics $h_i$. The <partition of unity> theorem for a Hausdorff second-countable smooth manifold supplies nonnegative smooth functions $\rho_i$, with locally finite supports contained in the respective cover members, such that $\sum_i\rho_i=1$. Set
$$
h=\sum_i\rho_i h_i,
$$
extending each weighted metric by zero off its chart. Local finiteness makes this smooth. For every nonzero fiber vector $v$, at least one positive weight gives $h(v,v)=\sum_i\rho_i h_i(v,v)>0$. Thus $h$ is a <fiber metric> and provides an <orthogonal structure>.
One definition of an <orthogonal connection> is that its <covariant derivative> preserves the <fiber metric>:
$$
X\bigl(h(s,t)\bigr)=h(\nabla_Xs,t)+h(s,\nabla_Xt)
$$
for all <vector fields> $X$ and smooth <sections of a vector bundle> $s,t$. Another is that its <parallel transport> along every smooth curve is an <isometry> between the endpoint fibers.
For the first implication, take parallel sections $s,t$ along a curve. Their covariant derivatives vanish, so the displayed identity makes $h(s,t)$ constant: <parallel transport preserves a fibre metric>. Conversely, if parallel transport preserves the <inner product>, choose parallel extensions of arbitrary initial fiber vectors along a curve with prescribed initial tangent $X$. Differentiating their constant inner product gives $(\nabla_Xh)(s,t)=0$ at the initial point. Since $X,s,t$ were arbitrary, the connection is <metric-compatible>.
The equivalent local description is that the <connection matrix in an orthonormal frame is skew-symmetric>. Writing $\nabla e_j=\sum_i A^i{}_j e_i$ in an <orthogonal trivialization>, differentiate $h(e_i,e_j)=\delta_{ij}$ to obtain
$$
\boxed{A^i{}_j+A^j{}_i=0,\qquad A^T=-A}.
$$
Conversely this matrix identity gives the metric-preservation formula by expanding arbitrary sections. It is the <orthogonal group> connection description of the same condition.
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