Let $S^1$ be the circle with circumference $1$. For each $0<r<1$, let $\mathcal{B}_r$ be the collection of open balls of diameter $r$ in $S^1$ (a.k.a. open arcs of length $r$ along $S^1$).
If $r\ne s$, can there exist a bijection $f: S^1\rightarrow S^1$ which induces a bijection $\mathcal{B}_r \rightarrow \mathcal{B}_s$ (via the canonical map $B\rightarrow f(B)$ for $B\in \mathcal{B}_r$)?
If $r$ and $s$ differ sufficiently, then (assuming $r>s$) there exists a subcover of $\mathcal{B}_r$ with fewer elements than any subcover of $\mathcal{B}_s$, hence the desired bijection cannot exist. For example, if $r={2\over 3}$, and $s = {1\over 3}$, then it is possible to cover $S^1$ with two elements of $B_r$ (take, for example, the two arcs of diameter ${2\over 3}$ centered around two antipodal points on $S^1$); however, $S^1$ cannot be covered with two elements of $B_s$ since the maximal length of a union of two elements of $B_s$ is ${2\over 3}$. Any bijection $f$ satisfying the above properties would need to send covers to covers (of the same cardinality), hence in this case no such bijection can exist.
But what if $r$ and $s$ do not differ sufficiently? Can such a bijection never exist if $r\ne s$, or can it exist if $r$ and $s$ are sufficiently close together?