cartesian product $A^2 = A$, possible? Do there exist non-empty sets $A$ such that $A\times A = A$?
$A\times A = A$ looks a little strange to me, since $A\times A$ seems somehow more complicated than $A$, hence it is unlike that they are equal, but then, I cannot think of any reason why there should not be such a (non-empty) set.

I assume you define $(a,b):=\{\{a\},\{a,b\}\}$; for other definitions of "ordered pair" the following proof can be adjusted.
Let $A$ be a set such that $A\times A=A$.
So for $x\in A$ there exist $y,z\in A$ such that $x=\{\{y\},\{y,z\}\}$. Especially, there exists a unique $y\in A$ such that $\{y\}\in x$. This defines a map $f\colon A\to A$ with $\{f(x)\}\in x$ for all $x\in A$.
Assume $A\ne 0$ and let $a\in A$ be an element. Then the sequence $a, f(a), f(f(a)), \ldots$ gives us a nonempty set $$B=\{\,f^n(a)\mid n\in\mathbb N_0\,\}\cup \{\,\{f^n(a)\}\mid n\in\mathbb N_0\,\}.$$
By the Axiom of Foundation, there exists $b\in B$ with $b\cap B=\emptyset$.
But if $b=f^n(a)$ then $\{f^{n+1}(a)\}\in b\cap B$; and if $b=\{f^n(a)\}$ then $f^n(a)\in b\cap B$. Contradiction! Therefore $A=\emptyset$.
