= Solution
\b[The printed assertion is false when maximal means largest in the natural-number labelling.] Both the original PDF and the TeX have this wording. The specified operation gives the <complete graph>, not the <Rado graph>.
Fix any two distinct vertices $u,v\in\mathbb N$, and choose $M\geq\max\{u,v\}$. Apply the <hypergraph extension property> of $G^{(3)}_{\mathrm{univ}}$ to $S=\{1,\ldots,M\}$, prescribing that the pair $\{u,v\}$ extend to a <hyperedge>; prescribe all other pairs in $S$ arbitrarily. The resulting witness $w\notin S$ has $w>M$, so $\{u,v,w\}$ is a <hyperedge> whose largest vertex is $w$. Deleting that vertex produces the <edge> $\{u,v\}$ in $G_0$.
Since $u,v$ were arbitrary,
$$
\boxed{G_0=K_{\mathbb N}.}
$$
The <Rado graph> has nonedges: its <hypergraph extension property> supplies a new vertex nonadjacent to any specified existing vertex. Therefore $K_{\mathbb N}$ and $G^{(2)}_{\mathrm{univ}}$ cannot be <isomorphic graphs>. This argument holds for every representative of the universal three-uniform structure labelled by $\mathbb N$, so changing its labelling does not repair the assertion. Equivalently, in an independently sampled representative each pair has infinitely many larger vertices that complete it to a triple, <almost surely>.
A valid related construction is to fix a vertex $a$ and take its <link hypergraph>: make $\{u,v\}$ an <edge> exactly when $\{a,u,v\}$ is a <hyperedge>, on the remaining <vertex set>. This is indeed the <Rado graph>. To prove it, take any disjoint finite sets $U,W$ of vertices other than $a$, put $S=\{a\}\cup U\cup W$, and prescribe triples through the pairs $\{a,u\}$ for $u\in U$ and no triples through $\{a,w\}$ for $w\in W$. Complete all other pair prescriptions arbitrarily. The <hypergraph extension property> produces a new vertex adjacent in the link to all of $U$ and none of $W$. Uniqueness from part (i), for $r=2$, now proves the claimed <hypergraph isomorphism> for this corrected operation.
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