For any set $X$ let $[X]^2 = \big\{\{a,b\}: a\neq b\in X\big\}$. We say that a graph $G$ is *self-complementary* if $G\cong \bar{G}$ where $\bar{G} = (V, [V]^2\setminus E)$. 


Given an infinite cardinal $\kappa$, is there a collection ${\cal C}$ of pairwise non-isomorphic self-complementary graphs on the vertex set $\kappa$ such that ${\cal C} = 2^\kappa$?



Collection of pairwise non-isomorphic infinite self-complementary graphs
As you know, there are $2^\kappa $ nonisomorphic graphs of cardinality $\kappa$ for every infinite cardinal $\kappa$. (In fact there are $2^\kappa$ nonisomorphic *trees* of cardinality $\kappa$, see [this answer](https://mathoverflow.net/questions/281039/mutually-non-isomorphic-connected-graphs-on-kappa-points/358126#358126).) I will show how to turn them into nonisomorphic self-complementary graphs of the same cardinality.


Given a graph $G$ of cardinality $\kappa$, let $G\_1,G\_2$ be two copies of $G$, and let $H\_1,H\_2$ be two copies of the complement of $G$, and add edges joining all vertices of $G\_1$ to all vertices of $H\_1$, all vertices of $H\_1$ to all vertices of $H\_2$, and all vertices of $H\_2$ to all vertices of $G\_2$. (In other words, we take the self-complementary graph $P\_4$ and replace the end vertices with copies of $G$, the internal vertices with copies of $\overline G$.) In this way we get a self-complementary graph $S$ of cardinality $\kappa$.


To recover $G$ from $S$, choose a vertex $y\_0$ of eccentricity $3$ and let $X$ be the set of all vertices $x$ such that $d(x,y\_0)=3$; then the subgraph of $S$ induced by $X$ is a copy of $G$.


