How big is the proper class of all sets? Let $\operatorname{ZFC}^{-}$ be the theory of $\operatorname{ZFC}$ minus the axiom of foundation and define the proper classes $G$ and $V$ as follows: 
$G:=$ The proper class of all sets.
$V:=$ The proper class of Von neumann cumulative heirachy.
And let the statements $G=V$ and $|G|=|V|$ be:
$G=V~:~~\forall x~\exists y~(\operatorname{ord}(y)~ \wedge "x\in V_{y}")$ 
$|G|=|V|~:~~\exists F\colon G \longrightarrow V~~$ a one to one function.
Using the axiom of foundation it is clear that we have $G=V$ and $|G|=|V|$ means that $G$ is as "small" as Von Neumann's cumulative hierarchy.
Now the question is: "How big is the proper class of all sets in the absence of the axiom of foundation? " 
In the other words which one of the following statements are true?
(1) $\operatorname{Con}(\operatorname{ZFC}) \longrightarrow \operatorname{Con}(\operatorname{ZFC}^{-} + G\neq V)$
(2) $\operatorname{Con}(\operatorname{ZFC}) \longrightarrow \operatorname{Con}(\operatorname{ZFC}^{-} + |G|\neq|V|)$ 
 A: Both the statement are true, which is fine because we talk about relative consistency here.
If $\sf ZFC$ is consistent then we know how to generate a model of $\sf ZFC^-+\it G\neq V$.
On the other hand, it is consistent that we have a proper class of atoms: sets of the form $x=\{x\}$, with global choice. Then by taking a class sized permutation model we can ensure that the class of the atoms is not well-orderable, while the axiom of choice for sets holds, and global choice for well-founded sets holds.
In that case we have $G\models\sf ZFC^-$ but $|G|\neq|V|$.
A: It seems that most of the usual anti-foundational set theories, when combined with the axiom of choice, imply that $|V|=|G|$. For example, in the case of Aczel's anti-foundation axiom AFA, we have
Theorem. ZFC-foundation + AFA proves $|V|=|G|$. 
Proof. Since $V$ is contained in $G$, it suffices to find an injection of $G$ into $V$. Consider any set $x$, not necessarily well-founded. Let $y$ be the transitive closure of $\{x\}$, so $y$ has $x$ and all the hereditarily elements of $x$, and let $\langle y,{\in},x\rangle$ be the corresponding accessible pointed graph, using the $\in$ relation. By AC, this graph has isomorphic copies $\langle h,\to,a\rangle$ in $V$, since we may well-order the nodes and thus find an isomorphic copy built on ordinals. Let $F(x)$ be the collection of all such graphs isomorphic to $\langle y,{\in},x\rangle$ chosen of $\in$-minimal rank in $V$. This is a set in $V$, and furthermore, $F$ is injective, since we can recover $y$ and hence $x$ from any graph isomorphic to $\langle y,{\in},x\rangle$. So we have injections both ways $V\to G$ and $G\to V$, and so they are bijective. QED
The same argument works with many of the other anti-foundational set theories, provided that equality of sets is determined by properties of the underlying $\in$-graph on the hereditary closure of the set. One prominent exception to this is that the Boffa AFA does not have this property.
