Is the differential of the flow the identity near regular points? For a vector field $X$ on a manifold $M$ we have, at least locally and for short time, a flow $\psi_t$ of $X$. If $X$ is regular at some point, we can find coordinates rectifying the vector field such that $\partial_1=X$. Then the representation of $\psi_t$ is just $(x_1+t,\dots,x_n)$. But the representation of the differential $d \psi_t : T_pM\to T_{\psi_t(p)}M$ is the Jacobian matrix of this translation, i.e. the identity. But both the differential of a map and the identity are invariantly defined, so if they are equal in some coordinate representation, they must be equal. In other words, $d\psi_t$ is the identity at every regular point for small enough $t$. I would have expected more complicated behavior from the differential, so what did I do wrong?
 A: It seems the confusion would be resolved if things are written out more explicitly.
It might be better to say that the flow box theorem gives a smooth conjugacy between $\psi$ and the translation flow, that is, for $p\in M$ with $X(p)\neq 0$, there is an open set $U$ containing $p$ and a $C^r$ diffeomorphism $\Phi:U\to B^1\times B^{\dim(M)-1}$ such that $\Phi\circ \psi_t(x)=\Phi(x)+te_1$ (when applicable), where $B^i$ is an open ball of dimension $i$, and $C^r$ is the regularity of the vector field $X$. The local translation flow $\Phi\circ \psi_t\circ \Phi^{-1}$ does indeed have derivative identity, but not necessarily $\psi_t$ itself. Indeed, for $p$ a regular point $\psi_t(p)\neq p$, so there is no natural way to consider the vector spaces $T_pM$ and $T_{\psi_t(p)}M$ as identical. (Note that we also haven't specified a Riemannian metric or a connection.)
What is true is that $T_p\psi_t:T_pM\to T_{\psi_t(p)}M$ can be represented as the identity matrix w/r/t certain bases, which means in invariant language that it is an isomorphism of vector spaces.

As a sanity check consider $M=\mathbb{R}$ and $X:x\mapsto cx$ with $c\neq0$. Then the associated global flow is $\psi:(t,x)\mapsto \psi_t(x)=e^{ct}x$. Consider $\Phi(e^{ct}x)=\Phi(x)+t$, we know by the flow box theorem that this equation has a solution $\Phi$ that is at least locally a diffeomorphism. Taking derivatives w/r/t $t$ and evaluating at $t=0$ one obtains, for $x\neq0$,
$$\Phi'(e^{ct}x)ce^{ct}x=1 \implies \Phi'(x)cx=1 \implies \Phi(x)=\ln(|x|)/c+C.$$
Thus the flow $\psi$ has exponential growth intrinsically, however it can be considered simply as translations in logarithmic coordinates (and only in such coordinates can it be considered to have "derivative identity").
