Existence of normal orthogonal frame on sphere such that it is normal at every point in a neighbourhood? It seems that on sphere $S^{n-1}$, there exists a better frame than the usually normal frame.
In the literature, some author asserts that there exists a local orthogonal frame $\left\{e_i\right\}_{i=1}^{n-1}$ on $S^{n-1}$, such that
$$
 [\bar\nabla_{e_i}e_j](x)=-x\delta_{ij},\quad\forall x\in U,
$$
where $\bar\nabla$ is the Levi-Civita connection of $R^n$ (which is just the directional derivative), and $x$ the the base point on $S^{n-1}$, $U$ is a neighbourhood.
I try to figure this out, but it seems not as simple as I think.
 A: This assertion is not true. 
If $\{e_i\}$ is such a frame, then the Gauss formula yields
$$
-x\delta_{ij} = (\overline \nabla _{e_i} e_j)(x) = (\nabla _{e_i} e_j)(x) + II(e_i,e_j)(x),\tag{$*$}
$$
where $\nabla$ is the intrinsic Levi-Civita connection on $S^{n-1}$ and $II$ is the (vector-valued) second fundamental form.
Note that $-x$ is the inward-pointing unit normal vector at each $x\in S^{n-1}$.    Decomposing $(*)$ into its normal and tangential components yields
\begin{align*}
-x\delta_{ij} &= II(e_i,e_j)(x),\\
0 &= (\nabla _{e_i} e_j)(x).
\end{align*}
EDIT: Fixed the second equation above.
A standard computation shows that $II(X,Y)(x) = -\left<X,Y\right>x$ for any vector fields $X$ and $Y$ tangent to the unit sphere, so the first equation above shows that $\{e_i\}$ is an orthonormal frame. The second equation shows that each vector field $e_j$ is parallel on $S^{n-1}$. However, the existence of a parallel orthonormal frame would imply that $S^{n-1}$ is flat, which it certainly is not.
[BTW, your title seems inappropriate. The assertion you quoted doesn't have anything to do with normal coordinates, or in fact with any coordinates at all.]
EDIT 2: The revised question (posed in a comment below) is to find a local orthogonal frame $\{e_i\}$ satisfying 
$$
 (\overline\nabla_{e_i}e_j)(x)=-xf(x)\delta_{ij},
$$
where $f\colon S^{n-1}\to\mathbb R$ is a conformal factor.
This is also impossible. In fact, more generally, it's impossible to find any local frame for which $\overline\nabla_{e_i}e_j$ is always a multiple of $x$.  Such a frame would satisfy
$$
 (\overline\nabla_{e_i}e_j)(x)=-x g_{ij}(x),
$$
for some smooth functions $g_{ij}$ on $U$.
If $\{e_i\}$ is such a frame, then the analysis above shows that the frame is parallel, and 
$\left< e_i, e_j\right> = g_{ij}(x)$.
The fact that the Levi-Civita connection is compatible with the metric implies
\begin{align*}
e_k g_{ij}(x) = e_k(\left< e_i, e_j\right>) = \left< \nabla_{e_k} e_i, e_j\right> + \left< e_i, \nabla_{e_k}e_j\right>  = 0.
\end{align*}
Since this is true for all $i,j,k$, it implies that each coefficient $g_{ij}$ is constant. Thus we can make a constant-coefficient change of frame (using Gram-Schmidt) to construct an orthonormal frame, each vector of which is still parallel, and then we're back to the previous situation.
