Why is a diffeomorphism preserving a parallelism locally uniquely determined by its value at $1$ point?

Question

According to the title, why is a diffeomorphism preserving a parallelism locally uniquely determined by its value at $1$ point?

Answer 1

Suppose $f,g$ are two diffeomorphisms, with the same values on $x$, that preserve the parallelization. Consider $h = gf^{-1}: M \to M$; this sends $x$ to itself. We want to show this is the identity. First, the set of points where $h$ is the identity is closed (because $h$ is continuous). Second, we show this is open. Work in a chart that respects the parallelization: one such that $e_i = \partial/\partial x_i$ at every point, and we may as well demand that $h(x) = 0 = x$. Then saying that $h$ preserves the parallelization is the same as saying that $\partial h/\partial x_i = 1$ for all $i$, and taking integrals we see that $h(x) = x$ for all $x$. So $h$ is the identity on this chart, and hence the set of points for which $h$ is the identity is both open and closed, hence everything.