I am wondering if any smooth bijection $f \in C^\infty(\mathbb{C})$ on $\mathbb{C}$ preserves the Hausdorff dimension of any given subset $A \subset \mathbb{C}$? In particular, I am working on understanding a paper where the author claims that the inverse of the Cayley transform given by $$ \varphi(z)=\frac{iz+1}{z+i}, \quad z \in \mathbb{C}$$ preserves the Hausdorff dimension of any given subset $A \subset \left\lbrace z \in \mathbb{C} \ | \ |z|=1\right\rbrace $ because $\varphi$ is smooth. I already know that a homeomorphism does not always preserve the Hausdorff dimension and that bi-Lipschitz bijections do preserve the Hausdorff dimension. I noticed that $\varphi^{-1}$ is bi-Lipschitz on $\mathbb{C} \setminus B(i,\epsilon)$ for any $\epsilon >0$ but I don't know if this is a sufficient condition if there is an accumulation point at $i=\varphi(\infty)$ Is the author right?
Any input is helpful! A counter example or proof would be awesome!