# Set of all sets those are equipotent to a given set?

Two sets are said to be 'equipotent' if there is a bijection between them. For a given set $A,$ consider the class $\Bbb{A}$ of all sets those are equipotent with $A.$ Is $\Bbb{A}$ form a set?

My answer is "No" unless $A=\emptyset.$ In order to prove this, my idea is to use the fact that class of all singleton sets is not a set.

1) Is my conclusion correct?
2) Is there any better (direct) way to prove this?

There are several ways to prove it, and I'm not entirely sure which one you have in mind based on your idea, but here's a hint that follows that idea: what is the cardinality of $\{x\}\times A$, for any set $x$?
• My idea is, if $A$ is non-empty, then $(A\setminus\{a\})\cup\{x\}$ is equipotent with $A$ for any singleton $\{x\}$ and some $a\in A.$ But your idea is way clear than it. – Bumblebee Jun 21 '17 at 5:27