Suppose $X$ is a non empty set and let $d_1, d_2$ be two metric on $X$, so we have two metric spaces over the same set $X$: $(X,d_1), (X,d_2).$
Suppose that there exists a surjective bi-Lipschitz map $\phi: (X,d_1)\to (X,d_2)$. Can we conclude that $d_1$ and $d_2$ are equivalent?
You did not specify what definition of equivalence you are using, but let me point out an example.
Take $X=\mathbb{R}$ with its usual metric $d_1(x,y)=|x-y|$. Take $\phi : X \to X$ to be a horrible, awful, very bad bijection. Do your worst. Make it discontuous everywhere, etc. etc. Define $d_2(x,y) = d_1(\phi(x),\phi(y))$. Then I see no reasonable definition of equivalence of metrics under which one would say that $d_1$ and $d_2$ are equivalent.
ADDED: With the definition of equivalence given by the OP in comments, the metrics $d_1$ and $d_2$ are not equivalent, indeed the identity map is not even continuous.