I have no idea what to do here.Show that the set of all isomorphism types of relations on $\displaystyle X$ has cardinality less than $\displaystyle 2^{2^{|X|^2}}$.

From the previous question, I know that if $\displaystyle B(X)$ denotes all the bijections $\displaystyle X \rightarrow X$, then $\displaystyle |X| \leq |B(X)| \leq |X|^{|X|}$.

I also know that if $\displaystyle |X|^2=|X|>1$ then $\displaystyle |B(X)|=2^{|X|}$.

I'm not sure if the previous question applies to this one. If it does, then I think I need to show that $\displaystyle |X|=2^{|X|^2}$. Unfortunately, this doesn't satisfy the $\displaystyle |X|^2=|X|$ condition. I also can't see why this would be the case.

Does anyone see how to do this question?