Recall the Inverse Function Theorem say
If $f: X \to Y$ is smooth, then if $df_x$ is an isomorphism at $x$, then $f$ is a local diffeomorphism at $x$.
Now this also means that $f$ being a local diffeomorphism is equivalent to the identity map $I$. That is, there are local parametrization $\psi: U \to X$ and $\phi: U \to Y$ such that $$f = \psi^{-1} \circ I \circ \phi = \phi(\psi^{-1}).$$
Q1: This should be obvious, but why is identity always equivalent to $f$? Doesn't this mean that every smooth map with an isomorphic derivative at the same point $x$ is equivalent to each other? Because if we have two different maps $$f = \psi^{-1} \circ I \circ \phi $$ and $$\bar{f} = \alpha^{-1}\circ I \circ \beta,$$ then $$f= \psi^{-1} \circ \beta^{-1} \circ \bar{f} \circ \alpha \circ \phi$$
This last part ties in with my other question. (sorry the open sets are different, just answer the first part).
If $f$ is a local diffeo at $x$, then let $\phi:U\to M$ be a local parametrization of a neighborhood $\phi(U)$ of $x$ such that $\phi(0) = x$. If $f$ maps a neighborhood $V$ of $x$ diffeomorphically onto $f(V)\subset N$, then by replacing $U$ with $\phi^{-1}(V\cap \phi(U))$, we get a diffeo $\psi:U\to N$ defined by $\psi = f\circ \phi$ such that $\psi(0) = f(\phi(0)) = f(x)$. Then, we have that $f$ is equivalent to the identity $I:\Bbb R^n\to\Bbb R^n$ via $$ f = (f\circ\phi)\circ I \circ \phi^{-1} = \psi\circ I\circ \phi^{-1}. $$