It is a well known fact that given a Riemannian manifold $M$, then for all $x\in M$, there exists a local orthonormal frame $\{e_i\}$ on an open neighborhood of $x$ such that:
$$(\nabla e_i)(x)=0$$ for all $i$.
Suppose $E\rightarrow M$ is a vector bundle over $M$, and $\nabla$ is a covariant derivative on $E$. Does the same statement hold?
I think the answer is yes, but I am not quite sure how to prove it. Perhaps one could write the covariant derivative as connection one form on the bundle of linear frames of $E$, and then make a parallel transport argument, but I can't quite see where to really start. Perhaps I am wrong though, and the answer is no. If so, how can I construct a counter example?
This is actually easy, per Ted Shiffrin's comment, and I believe works in particular for metric connections as well. Just restrict the following answer to the bundle of orthonormal frames for your vector bundle and everything should as earlier.
Let $e_i$ be a local frame on $E_U$, and $x\in M$ be a point. Suppose that $f_i$ is any other local frame related to $e_i$ by a matrix of functions $A\in GL(V)$:
$f_i=A^j_ie_j$.
If, in the original local frame we have:
$$\nabla e_i = \omega_i^je_j$$
for the connection one form $\omega$, then we see that by Leibniz property of covariant derivatives:
$$\nabla f_i= (dA_i^je_j+A^j_k \omega^k_i)e_j$$
It follows that for this to be zero at a point we need:
$$(dA)_x=-(A\omega)_x$$
Choose a coordinate chart $\phi$, shrinking $U$ if necessary, such that $\phi(x)=0\in \mathbb{R}^n$, and let $x^\mu$ be the coordinates for this chart. Let $\omega_\mu=\omega(\partial_\mu)\in\mathfrak{gl}(V)$. Shrink $U$ further such that the function:
$$A=I-x^\mu \omega_\mu(0)$$
lies in $GL(V)$. Then we see that:
$$dA=-\omega(0)$$
It follows that $A$ takes $e_i$ to a frame which is then parallel at the point the $x\in M$, as desired.