Complete metric spaces question

Question

If two metric spaces let's say $(X,d)$ and $(X,h)$ are complete, does it imply that $h$ and $d$ are topologically equivalent ?

Answer 1

Of course not, let $X =\mathbb{R}$ and $d(x,y)=|x-y|$ the usual metric and $h(x,y) = 0 (x=y), 1 (x \neq y)$ the discrete metric. Both are complete, but they are very non-equivalent.