Aut(U_Q^<) is extremely amenable

The age of $U_{\mathbb{Q}}^{<}$ is the class of finite ordered rational metric spaces, and $U_{\mathbb{Q}}^{<}$ is its Fraïssé limit, since it is countable, universal and homogeneous for that class by Corson's ordered-rational permutation model. Encode rational distances by a binary relation for each rational value, together with the order relation; this is a countable relational language. Its automorphism group is closed in the permutation group of the underlying countable set: failure to preserve a relation is witnessed by a finite tuple and remains a failure on a basic neighbourhood.

Every finite ordered rational metric space is rigid by [L1], and the age is a Ramsey class by [F1]; hence the hypotheses of the KPT criterion [F2] hold for the Fraïssé limit $U_{\mathbb{Q}}^{<}$, and $\mathrm{Aut}(U_{\mathbb{Q}}^{<})$ is extremely amenable.

Put $G:=\mathrm{Aut}(U_{\mathbb{Q}}^{<})$ and $H:=\mathrm{fix}(E)$. In the pointwise-convergence topology $H$ is an open subgroup of $G$, since fixing the finitely many points of $E$ is a basic identity neighbourhood.

Let $X$ be a nonempty compact Hausdorff $H$-flow. Inside the product $X^G$, define the coinduced space $Y:=\{\mathrm{\Phi }:G\to X:\mathrm{\Phi }(hg)=h\cdot \mathrm{\Phi }(g)\text{ for all }h\in H,g\in G\}\mathrm{.}$ It is nonempty: AC chooses one representative of every left $H$-orbit in $G$; fix one $x_0\in X$, assign that same value at every representative, and then the displayed rule extends it uniquely to that orbit.

The space $Y$ is closed in $X^G$: for fixed $h,g$, the equation $\mathrm{\Phi }(hg)=h\cdot \mathrm{\Phi }(g)$ is an equaliser of two continuous coordinate maps and is closed because $X$ is Hausdorff. Arbitrary intersections of these closed equalisers are closed. Hence [F3] makes $Y$ compact with its subspace topology. It is Hausdorff: two distinct functions differ at some coordinate, where disjoint open neighbourhoods in $X$ pull back to disjoint cylinder neighbourhoods in $Y$.

Define a $G$-action on $Y$ by $(a\cdot \mathrm{\Phi })(g):=\mathrm{\Phi }(ga)\mathrm{.}$ The defining equivariance of $Y$ is preserved, since $\mathrm{\Phi }(hga)=h\cdot \mathrm{\Phi }(ga)$. It is a left action: $a\cdot (b\cdot \mathrm{\Phi })$ evaluated at $g$ is $\mathrm{\Phi }((ga)b)=\mathrm{\Phi }(g(ab))$, and the identity acts trivially. This action is continuous. Indeed, at $a_0\in G$ and for each of finitely many output coordinates $g_i$, openness of $H$ gives a neighbourhood on which $h_i(a):=g_{i}a{a}_0^{-1}{g}_i^{-1}\in H$; then $g_{i}a=h_i(a)g_i{a}_0$ and $\mathrm{\Phi }(g_{i}a)=h_i(a)\cdot \mathrm{\Phi }(g_i{a}_0)$. Continuity of $h_i$, of the $H$-action, and the product topology at the finitely many fixed coordinates $g_i{a}_0$ therefore give joint continuity.

Extreme amenability of $G$ from [step 2.1] gives a $G$-fixed $\mathrm{\Phi }\in Y$. The right-translation action then makes $\mathrm{\Phi }$ constant, since $\mathrm{\Phi }(a)=(a\cdot \mathrm{\Phi })(1)=\mathrm{\Phi }(1)$ for every $a\in G$. For $h\in H$, the defining equation for $Y$ gives $h\cdot \mathrm{\Phi }(1)=\mathrm{\Phi }(h)=\mathrm{\Phi }(1)$, so $\mathrm{\Phi }(1)$ is an $H$-fixed point of $X$.

Thus every nonempty compact Hausdorff $H$-flow has a fixed point, so $H=\mathrm{fix}(E)$ is extremely amenable. Since $E$ was arbitrary and $E=\mathrm{\varnothing }$ gives the whole group, the stated group and all required finite point stabilisers are extremely amenable.