Let $f, f_n: \mathbb{R}^m \rightarrow \mathbb{R}^m$ such that $f^{-1}, f_n^{-1}$ exist and are globally Lipschitz continuous for all $n$ (same constant $L$).
Assume that $f_n := f + \frac{1}{n} Id$. Then $\left( f_n \right)^{-1} \rightarrow f^{-1}$ uniformly on every compact set.
Prove or disprove that $\left\| \left( f_n \right)^{-1} - \left( f \right)^{-1}\right\| \leq \frac{c}{n}$, for some $c>0$.
Comments: $\left( f_n \right)^{-1} \rightarrow f^{-1}$ follows from the Lipschitz continuity property, see: Convergence of sequences of inverse functions. However, that does not exploit the fact that $\left( f_n \right)^{-1} = \left( f + \frac{1}{n} Id \right)^{-1}$.
Attempt: If we take $x = \left( f + \frac{1}{n}Id \right) (y) = f(y) + \frac{1}{n} y$, then $\left\| \left( f + \frac{1}{n}Id \right)^{-1}(x) - f^{-1}(x) \right\| = \left\| y - f^{-1}\left( f(y) + \frac{1}{n} y \right) \right\| = \left\| f^{-1}\left( f(y) \right) - f^{-1}\left( f(y) + \frac{1}{n} y \right) \right\| \leq L \left\| f(y) - f(y) + \frac{1}{n} y \right\| = L \left\| \frac{1}{n} y \right\| = \frac{L}{n} \left\| y \right\|$.
So the statement holds if $\left\| y \right\|$ is bounded, i.e. if the domain of $f$ is bounded. Otherwise it seems the statement may be false.
Notice that $f_n$ is bijective for sufficiently large $n$.
Big hint: $$ \| f_n^{-1} - f^{-1} \| = \| \operatorname{id} - f^{-1} \circ f_n \| = \| f^{-1}\circ f - f^{-1} \circ f_n \|. $$