Consider the two circles $D$ and $E$ such that $d(D,E)$ is maximal. Then $d(f(D),f(E))=d(D,E)$, and either $f(D)=D, f(E)=E$ or $f(D)=E, f(E)=D$. Now assume for two circles $A,B$, $d(f(A),f(B))>d(A,B)$. Then $d(f(D),f(E)) = d(D,E) \leq d(D,A)+d(A,B)+d(B,E) <$ $\min(d(f(D),f(A))+d(f(A),f(B))+d(f(B),f(E)),$ $d(f(D),f(B))+d(f(B),f(A))+d(f(A),f(E))) = d(f(D),f(E))$. Absurd.

julien_santini wrote: Consider the two circles $D$ and $E$ such that $d(D,E)$ is maximal. Then $d(f(D),f(E))=d(D,E)$, and either $f(D)=D, f(E)=E$ or $f(D)=E, f(E)=D$. Now assume for two circles $A,B$, $d(f(A),f(B))>d(A,B)$. Then $d(f(D),f(E)) = d(D,E) \leq d(D,A)+d(A,B)+d(B,E) <$ $\min(d(f(D),f(A))+d(f(A),f(B))+d(f(B),f(E)),$ $d(f(D),f(B))+d(f(B),f(A))+d(f(A),f(E))) = d(f(D),f(E))$. Absurd. I don't think this argument works unless you can show that $f$ is continuous. Anyway, I think we just first show that the function $f$, restricted on the largest circle, is an isometry. Then, get rid of this circle, and proceed inward to the smallest circle.

(Due to Chung Ping Ngai, HK IMO team member) First off, f must be bijective, for otherwise there exists two distinct elements (say, X, Y) in S which are mapped to the same element, so that d(f(X), f(Y)) = 0 < d(X, Y), a contradiction. Note that since S is finite, for all A, B in S we have $f^p(A)=A$ and $f^q(B)=B$ for some positive integers p, q. Then $f^{pq}(A)=A$ and $f^{pq}(B)=B$. But $d(f^{pq}(A), f^{pq}(B)) \geq d(f^{pq-1}(A), f^{pq-1}(B)) \geq ... \geq d(A, B)$, so all $\geq$ signs must become equal signs. QED.

Ohhh. OK. I thought $S$ was the set of points of $n$ circles. Apologies to the first reply.