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 ${\cal C} = 2^\kappa$?
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.) 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 $G$.
