= Solution
Assume $X$ is nonempty, as is necessary for a <probability measure> to exist. Give its <isometry group> $I_X$ the uniform <metric>
$$
D(g,h)=\sup_{x\in X}d(gx,hx).
$$
Every 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 $d(g^mx_0,g^nx_0)=d(x_0,g^{n-m}x_0)\geq\delta$. 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 $D(gh,g'h')\leq D(g,g')+D(h,h')$ 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 \b[$I_X$ 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 $\varepsilon$-separated subsets of maximum <cardinality> in a <compact metric space>, join points whose distance is less than $\varepsilon$. If some $S\subseteq A$ had fewer than $|S|$ neighbours, replacing those neighbours in $B$ by $S$ would produce a larger $\varepsilon$-separated set. This contradicts maximal <cardinality>. Hence there is a bijection between $A$ and $B$ moving each point by less than $\varepsilon$.
Choose $\varepsilon_n\downarrow0$ and maximum-<cardinality> $\varepsilon_n$-separated sets $A_n\subset I_X$. Such sets exist: a finite covering by balls of radius less than $\varepsilon_n/2$ bounds every separated set's <cardinality>. Fix $x_0\in X$ and define the empirical <probability measures>
$$
\mu_n=\frac1{|A_n|}\sum_{a\in A_n}\delta_{ax_0}.
$$
For $f\in C(X)$ let $\omega_f(r)=\sup_{d(x,y)\leq r}|f(x)-f(y)|$, which tends to zero with $r$ by <uniform continuity>. For every $h\in I_X$, left invariance of $D$ makes $hA_n$ another maximum-<cardinality> separated set. Match it to $A_n$ as above. Then
$$
\left|\int f\,d\mu_n-\int f(hx)\,d\mu_n(x)\right|\leq\omega_f(\varepsilon_n).
$$
The 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 $\int f\,d\mu=\int f\circ h\,d\mu$. 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 $x=hx_0$, right invariance of $D$ lets us match $A_nh$ with $A_n$. Transitivity therefore gives the uniform estimate $|F_n(x)-F_n(x_0)|\leq\omega_f(\varepsilon_n)$ for every $x\in X$. If $\nu$ is any invariant <Borel probability measure>, then
$$
\int f\,d\nu=\int F_n\,d\nu,\qquad
\left|\int f\,d\nu-F_n(x_0)\right|\leq\omega_f(\varepsilon_n).
$$
Two invariant <probability measures> consequently differ on the integral of $f$ by at most $2\omega_f(\varepsilon_n)$, hence agree on every <continuous function>. <Continuous functions> determine <Borel probability measures> on a <compact metric space>, so \b[the invariant <probability measure> exists and is unique].
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