Disjoint union of copies of Henson graph is not homogenous

Question

From Wiki, the Henson graph $G_i$ is an undirected infinite graph, the unique countable homogeneous graph that does not contain an i-vertex clique but that does contain all $K_i$-free finite graphs as induced subgraph. A structure is homogenous if every isomorphism between finite substructures extends to an automorphism. I believe it can be shown that a Henson graph on its own is in fact homogenous - however, when taking the disjoint union of one with a copy of itself, this is not the case - I suspect we use the fact that we can form an isomorphism with edges not connected in one copy with vertices in the same copy, as well as edges trivially not connected in the other copy, and this causes problems. I am unsure how to proceed, though.

Answer 1

Let $a\in G_1$ and let $b,c\in G_2$ be non adjacent. The map $a,b\mapsto b,c$ does not extend to an automorphism.

In fact, in $G_2$ there is a vertex $d$ adjacent to both $b,c$. But in $G_1\cup G_2$ there is no vertex adjacent to both $a,b$.