 Solution

Assume $X$ is nonempty, as is necessary for a probability measure to exist. Give its isometry group $I_X$ the uniform metricEvery distance-preserving self-map of a compact metric space is onto. Indeed, if $x_0\notin gX$, then $\delta =d(x_0,gX)>0$, and for $n>m$ the iterates satisfy $d(g^m{x}_0,g^n{x}_0)=d(x_0,g^{n-m}{x}_0)\ge \delta$. This infinite separated set contradicts compactness. Consequently $I_X$ is a group under composition, with the identity and inverses again isometries.

The metric $D$ is invariant under both left and right multiplication: left invariance uses preservation of distances, and right invariance uses surjectivity. Inversion is an isometry for $D$, and $D(gh,g^{\prime }{h}^{\prime })\le D(g,g^{\prime })+D(h,h^{\prime })$ proves continuity of multiplication. Moreover, $I_X$ is compact. A sequence of isometries has a subsequence converging at every point of a countable dense subset of $X$, by a diagonal argument. Their common Lipschitz constant $1$ and finite epsilon-nets of $X$ make this subsequence uniformly Cauchy. Its uniform limit preserves distances and is onto by the preceding argument. Thus $I_X$ is a compact metrizable topological group. A transitive group action means that for every $x,y\in X$ some $g\in I_X$ has $gx=y$.

We construct the invariant Borel probability measure directly. The Hall marriage theorem says that a finite bipartite graph with equal-sized vertex classes $A,B$ has a perfect matching precisely when every subset $S\subseteq A$ has at least $|S|$ neighbours in $B$. Its following consequence is useful. If $A,B$ are $\epsilon$-separated subsets of maximum cardinality in a compact metric space, join points whose distance is less than $\epsilon$. If some $S\subseteq A$ had fewer than $|S|$ neighbours, replacing those neighbours in $B$ by $S$ would produce a larger $\epsilon$-separated set. This contradicts maximal cardinality. Hence there is a bijection between $A$ and $B$ moving each point by less than $\epsilon$.

Choose ${\epsilon }_n\downarrow 0$ and maximum-cardinality ${\epsilon }_n$-separated sets $A_n\subset I_X$. Such sets exist: a finite covering by balls of radius less than ${\epsilon }_n/2$ bounds every separated set's cardinality. Fix $x_0\in X$ and define the empirical probability measuresFor $f\in C(X)$ let ${\omega }_f(r)={\sup }_{d(x,y)\le r}|f(x)-f(y)|$, which tends to zero with $r$ by uniform continuity. For every $h\in I_X$, left invariance of $D$ makes $h{A}_n$ another maximum-cardinality separated set. Match it to $A_n$ as above. ThenThe allowed compactness of probability measures on a compact metric space supplies a subsequence converging to $\mu$ in the weak-star topology. Passing to that limit for each fixed $h,f$ gives $\int fd\mu =\int f\circ hd\mu$. The same subsequence works for every $h,f$, since the displayed estimate held for all of them before taking the limit.

For uniqueness, put $F_n(x)=|A_n{|}^{-1}{\sum }_{a\in A_n}f(ax)$. For $x=h{x}_0$, right invariance of $D$ lets us match $A_{n}h$ with $A_n$. Transitivity therefore gives the uniform estimate $|F_n(x)-F_n(x_0)|\le {\omega }_f({\epsilon }_n)$ for every $x\in X$. If $\nu$ is any invariant Borel probability measure, thenTwo invariant probability measures consequently differ on the integral of $f$ by at most $2{\omega }_f({\epsilon }_n)$, hence agree on every continuous function. Continuous functions determine Borel probability measures on a compact metric space, so the invariant probability measure exists and is unique.