Uncountably many countable graphs with no homomorphism between them

Question

By a graph I mean a pair $G = (V, E)$ where $V$ is a set and $E \subseteq \mathcal{P}_2(V) := \{\{a,b\}: a\neq b \in V\}$. A graph homomorphism between graphs $G, H$ is a map $f:V(G)\to V(H)$ such that $\{v, w\}\in E(G)$ implies $\{f(v), f(w)\} \in E(H)$.

If $G,H$ are graphs and there is a graph homomorphism $f:G\to H$ we write $G\to H$, and otherwise $G\not\to H$.

Let $C$ be the set of graphs such that $V(G)=\mathbb{N}$. We set $$E = \big\{\{G,H\}: (G,H\in C) \land (G\not\to H) \land (H\not\to G)\big\}.$$ Let $G_{\mathbb{N}} = (C,E)$.

Question: Does $G_{\mathbb{N}}$ have an uncountable clique?

Answer 1

Yes. Such cliques are called rigid families of graphs. For every infinite cardinal $\kappa$ there exists a rigid family of graphs of cardinality $2^\kappa$, such that each graph in the family has $\kappa$ vertices.

This is classical so I was surprised by the trouble of finding a neat reference tailored to your question. An overkill reference is Theorem 1 in Section 4 of P. Hell On some strongly rigid families of graphs and the full embedding they induce. Algebra Universalis 4 (1974), 108–126

Edit: Actually rigid families of graphs require a stronger condition - no nonidentity homomorphisms (including endomorphisms) between its members. But of course every rigid family satisfies your requirements.