If there is a bijection between $A$ and $\operatorname{Fin}(A)$, is there a bijection between $A$ and $A\times A$?

Question

If there is a bijection between $A$ and $\operatorname{Fin}(A)$ then there is a bijection between $A\times A$ and $A$?

Answer 1

The answer for (2) is indeed positive.

Recall that we can encode an ordered pair $(a,b)$ as the set $\{\{a\},\{a,b\}\}$. Therefore $A\times A$ is a subset of $\operatorname{Fin}(\operatorname{Fin}(A))$.

If $A$ and $\operatorname{Fin}(A)$ have the same cardinality, show that we actually get that $A$ and $\operatorname{Fin}(\operatorname{Fin}(A))$ also have a bijection between them, and conclude from the Cantor-Bernstein theorem the wanted result.

(Note that assuming the axiom of choice this is easier, since for a finite set, $A$, $\mathcal P(A)=\operatorname{Fin}(A)$ and therefore there is no bijection between them, and for every infinite set, $A\times A$ and $A$ have a bijection between them. But the crux of the question, I suppose, is without the axiom of choice.)