Consider the setup from here: Do these vector fields span an integrable distribution?
For any pair of points $p, q \in U$, show that there is a local diffeomorphism $F: U(p) \to U(q)$, such that $F_*(X_i) = X_i$ for $1 \le i \le n$.
I tried to solve it but couldn't figure out how to get started. Any hints would be appreciated.
Let $\overline{X}_i$, $\overline{U}$ be as in the setup from the linked question. The distribution $E$ is integrable by the linked question, so by the Frobenius Theorem, there exists an injective immersed submanifold $M \subseteq U \times \overline{U}$ near $(p, q)$ such that $T_{(a, b)}M = E_{(a, b)}$ for all $(a, b) \in M$. Let $\pi:U \times \overline{U} \to U$ be the projection map. Then $d\pi|_{(p, q)}$ sends $Y_i|_{(p, q)} = \left(X_i + \overline{X}_i\right)|_{(p, q)} = \left(X_i, \overline{X}_i\right)|_{p, q}$ to $X_i|_p$. The $Y_i|_{(p, q)}$'s and $X_i|_p$'s are linearly independent, which means the first $n$ columns of $\text{Jac}(\pi)$ form a nonsingular $n \times n$ matrix. By the Implicit Function Theorem, there exists smooth $F: U(p) \to U(q)$ such that $M = r(F)$. We have $F_*(X_i) = \overline{X}_i$ since $X_i + \overline{X}_i = Y_i$ is tangent to $M = r(F)$. So $F_*$ sends linearly independent $X_i|_p$'s to linearly independent $\overline{X}_i|_p$'s, which means it is nonsingular. By the Inverse Function Theorem, $f$ is a local diffeomorphism.