Orbits of points in compact metric space when there exists a point that is fixed by all isometries

Question

Let $(X,d)$ be a compact metric space and Iso$(X,d)$ be a set of all isometries $f : X \rightarrow X$.

For $x \in X$ orbit of $x$ is defined as: $$\operatorname{Orb}(x) := \{f(x) : f \in \operatorname{Iso}(X,d)\}.$$

Suppose $\operatorname{Iso}(X,d)$ is infinite and suppose there is a point $x_0 \in X$ such that $f(x_0) = x_0$, for every $f \in \operatorname{Iso}(X,d)$.

Let $x$ be a point in $X$ and define $r_x := d(x_0, x)$. Let $y \in \operatorname{Orb}(x)$ and $f \in \operatorname{Iso}(X,d)$ such that $f(x) = y$, we have $$r_x = d(x, x_0) = d(f(x), f(x_0)) = d(y, x_0).$$ It follows that $\operatorname{Orb}(x) \subseteq S(x_0, r_x)$, for every $x \in X$, where $S(x, r) := \{y \in X : d(x, y) = r\}$.

Is it maybe true that $\operatorname{Orb}(x) = S(x_0, r_x)$, for all $x \in X$, under these assumptions?

Answer 1

Consider the union of the unit disk in $\Bbb R^2\subset\Bbb R^3$ and the unit sphere in $\Bbb R^3$, centered at the origin (so the disk's boundary is an equator). The isometry group should be the orthogonal group $\mathrm{O}(2)$ (the $2$D reflections across lines achieved by $180^\circ$ $3$D rotations around them), with fixed point the origin, but the unit sphere is more than just one orbit.

Answer 2

There is a counterexample with only 4 points, let

$$ X= \{(0,0), (0,1), (1,0), (1,0.9)\} $$ endowed with the subspace metric from $\mathbb{R}^2$. (Draw this out, it is a slightly asymmetric square, almost like the unit square with its $(1,1)$ vertex slid down a bit)

Then the only isometry is the identity (the only contender for an isometry needs to permute the points $(0,0), (0,1)$ and $(1,0)$, by hand you can verify that none but the identity permutation can be isometrically completed) and thus the orbit of every element is the singleton containing itself, even though there are $2$ distinct points in $S((0,0),1)$.

We can turn this into a counterexample with an infinite isometry group: replace the vertex $(0,1)$ with a copy of the sphere $U$ with radius $\frac{1}{100}$ setting for every $y\in U$, and $x\in X\setminus \{(0,1)\}$

$$d(x,y) = d(x,(0,1))$$

Then any isometry effects only the elements of $U$, so $(1,0)\notin \mbox{Orb}(y)$ for every $y\in U$ even though $$(1,0) \in S((0,0),1) = S((0,0),d((0,0),y))$$

As an aside, when trying to answer this question, I convinced myself that there needs to be more than one element with a singleton orbit for a counterexample to work, perhaps you can try and see if this is sufficient.