Let $S$ be the unit circle and for any $x,y \in S$ let $d(x,y)$ be the lenght of the smallest arc between $x$ and $y$. A **bijective** map $\phi : S\longrightarrow S$ is called 1-iso if the following holds: 
$$ \forall x,y \in S \;\; d(x,y)=1 \longleftrightarrow d(\phi (x), \phi (y) ) =1$$
Now the question: Is there a subset $D$ of $S$ such that the only 1-iso map $\phi$ with the property that $\phi (D) = D$ and $\phi (S-D) = S-D$ is the identiry map ? 



On 1-iso maps and subsets of the unit circle
For any point $a\in S$, there is an infinite (in both directions) chain $\dots,a\_{-1},a\_0,a\_1,a\_2,\dots$ of points such that $d(a\_i,a\_{i+1})=1$, and no other points are at distance $1$ from the points of this chain. The whole circle is partitioned into (continuum of) such chains; for every chain we choose its representative $a\_0$.


Now find a bijection between the set of all chains and the set of all infinite strictly increasing sequences $(n\_1,n\_2,\dots)$ of positive integers. If a coset with representative $a\_0$ corresponds to a sequence $(n\_1,n\_2,\dots)$, we put into $D$ the elements $a\_0,a\_{n\_1},a\_{n\_1+n\_2},\dots$. Perform this operation for all chains.


Now, if $\phi$ is 1-iso and $\phi(D)=D$, then $\phi$ maps $\{a\_0,a\_{n\_1},a\_{n\_1+n\_2},\dots\}$ to either $\{b\_i,b\_{i+n\_1},b\_{i+n\_1+n\_2},\dots\}$ or $\{b\_i,b\_{i-n\_1},b\_{i-n\_1-n\_2},\dots\}$, and this image should equal to $D\cap (\dots,b\_{-1},b\_0,b\_1,\dots\}$. The second set cannot lie in $D$, while the first may equal $D\cap (\dots,b\_{-1},b\_0,b\_1,\dots\}$ only if $b\_0=a\_0$. Thus $\phi$ is the identity map.


