If $(X,d)$ is a metric space and $f:X \to X$ an injective map, I showed that $(X,d\circ f)$ is also a metric space. And if $f$ is an homeomorphism, then, $d$ and $d\circ f$ are topologically equivalent. But, what is a necessary and sufficient condition to get the lipschitz equivalence ?
My research : if there exists $c,C>0$ such that $cd(x,y) \leq d(f(x),f(y)) \leq Cd(x,y)$, then $f$ is C-lipschitz. So we need f to be lipschitz but it is sufficient ?
If you write the definition of Lipschitz equivalence, you should see rather easily that $d$ and $d\circ f$ are Lipschitz-equivalent if and only if $f$ is an homeomorphism such that $f$ and $f^{-1}$ are both Lipschitz.