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) $\operatorname{Con}(\operatorname{ZFC}) \longrightarrow \operatorname{Con}(\operatorname{ZFC}^{-} + |G|\neq|V|)$ 



How big is the proper class of all sets?
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 $|G|\neq|V|$.


