Let $A$ be a non-empty set, and $\mathcal{P}^*(A)$ denote the power set of $A$ excluding empty set. There is a natural equivalence relation on $\mathcal{P}^*(A)$:
for $S_1,S_2\in \mathcal{P}^*(A)$ (i.e. $S_1,S_2\subseteq A$), we say $S_1\sim S_2$ if there is a bijection between $S_1,S_2$.
The question I am concerning here is about the number of equivalence classes of $\mathcal{P}^*(A)$ under $\sim$. When $A$ is a finite set, then the number of equivalence classes is easily seen to be $|A|$; this can also be proved if $A$ is countable, by providing a bijection between $A$ and the equivalence classes of $\mathcal{P}^*(A)$.
Question: What is the number of equivalence classes of $\mathcal{P}^*(A)$ under $\sim$ if $A$ is uncountably infinite?
The number of equivalence classes in $\mathcal{P}^*(A)$ is intuitively seems to be $|A|$, but I am not getting any direction in case $A$ is uncountably infinite. I didn't find such thing (question/fact) in some standard texts of set theory.
It depends on what $A$ is. In fact, (assuming the axiom of choice) the number of equivalence classes of $\mathcal{P}^*(A)$ for $A$ uncountable can be any infinite cardinality at all, by choosing $A$ appropriately. By definition, if $|A|=\aleph_{\alpha}$, the infinite cardinals that are less than or equal to $|A|$ exactly the cardinals $\aleph_\beta$ for $\beta\leq\alpha$. So the number of different cardinalities of subsets of $A$ is $\aleph_0+|\alpha|$. Here $\alpha$ can be any ordinal at all, so $|\alpha|$ can be any cardinal.
Note that in particular, for instance, ZFC cannot answer this question for $A=\mathbb{R}$, since ZFC cannot determine which aleph number $|\mathbb{R}|$ is (it cannot even determine the cardinality of the ordinal $\alpha$ such that $\aleph_\alpha=|\mathbb{R}|$; all that can really be said is that $|\alpha|\leq|\mathbb{R}|$).