In some sense the empty set ($\emptyset$) and the global set of all sets ($G$) are the ends of the universe of the mathematical objects. The world which $ZFC$ describes has an end from the bottom and is endless from the top. Simply one can find an equiconsistent theory (respect to $ZFC$) which its world is endless from the bottom and bounded from the top by the set of all sets. It is sufficient to consider the theory $ZFC^{-1}$ ($ZFC$ inverse) which is obtained from $ZFC$ by replacing each phrase $x\in y$ in the axioms of $ZFC$ by the phrase $\neg (x\in y)$. This operation for example transforms the axiom of empty set of $ZFC$ to an statement which asserts "the set of all sets exists".

$[\exists x \forall y~~\neg(y\in x)]\mapsto [\exists x \forall y~~\neg \neg(y\in x)] $

Even the axiom of extensionality remains unchanged because we have:

$[\forall x\forall y~~(x=y\longleftrightarrow \forall z~~(z\in x\longleftrightarrow z\in y))]\mapsto [\forall x\forall y~~(x=y\longleftrightarrow \forall z~~(\neg (z\in x)\longleftrightarrow \neg (z\in y)))]$

So the "set of all sets" is unique in this theory. Even the equiconsistency simply follows from the fact that for all set (or proper class) $M$ and for all binary relation $E$ on it we have:

$\langle~M,E~\rangle \models ZFC \Longleftrightarrow \langle~M,M\times M\setminus E~\rangle \models ZFC^{-1}$

But it is trivial that $ZFC^{-1}\models \neg (\exists x \forall y~~\neg(y\in x))$ in the same way which one can prove $ZFC\models \neg (\exists x \forall y~~y\in x)$ by the Russell's paradox. But the situation seems rather strange when one wants to find an equiconsistent theory with $ZFC$ which has end points in both up and down direction because the existence of two contradictory objects like $\emptyset$ and $G$ seems ontologically incompatible in a same $ZFC$-like world. So the question is:

Question (1): Is there an $\mathcal{L}=\lbrace \in\rbrace$-theory $T$ such that the following conditions hold:

$(1)~Con(ZFC)\Longleftrightarrow Con(T)$

$(2)~T\models \exists !x~\forall y~~(y\in x)$

$(3)~T\models \exists !x~\forall y~~\neg (y\in x)$

Remark (1): Quine's new foundation axiomatic system ($NF$) is not an answer because its equiconsistency with $ZFC$ is still unknown.

Even one can define two dual sets from empty and global sets. The set which does not belong to any other set ($\emptyset^{\star}$) and the set which belongs to any set ($G^{\star}$).Now one can restate the question (1) as follows:

Question (2): Is there an $\mathcal{L}=\lbrace \in\rbrace$-theory $T$ such that the following conditions hold:

$(1)~Con(ZFC)\Longleftrightarrow Con(T)$

$(2)~T\models \exists !x~\forall y~~(x\in y)$

$(3)~T\models \exists !x~\forall y~~\neg (x\in y)$

In some sense the empty set ($\emptyset$) and the global set of all sets ($G$) are the ends of the universe of mathematical objects. The world which $ZFC$ describes has an end from the bottom and is endless from the top. Even in a straight forward way one can find an equiconsistent theory (respect to $ZFC$) which its world is endless from the bottom and bounded from the top by the set of all sets. It is sufficient to consider the theory $ZFC^{-1}$ ($ZFC$ inverse) which is obtained from $ZFC$ by replacing each phrase $x\in y$ in the axioms of $ZFC$ by the phrase $\neg (x\in y)$. This operation for example transforms the axiom of empty set of $ZFC$ to an statement which asserts "the set of all sets exists".

$[\exists x \forall y~~\neg(y\in x)]\mapsto [\exists x \forall y~~\neg \neg(y\in x)] $

Even the axiom of extensionality remains unchanged because we have:

$[\forall x\forall y~~(x=y\longleftrightarrow \forall z~~(z\in x\longleftrightarrow z\in y))]\mapsto [\forall x\forall y~~(x=y\longleftrightarrow \forall z~~(\neg (z\in x)\longleftrightarrow \neg (z\in y)))]$

So the "set of all sets" is unique in this theory. Even the equiconsistency simply follows from the fact that for all set (or proper class) $M$ and for all binary relation $E$ on it we have:

$\langle~M~,~E~\rangle \models ZFC \Longleftrightarrow \langle~M~,~M\times M\setminus E~\rangle \models ZFC^{-1}$

So it is trivial that $ZFC^{-1}\models \neg (\exists x \forall y~~\neg(y\in x))$ in the same way which one can prove $ZFC\models \neg (\exists x \forall y~~y\in x)$ by the Russell's paradox. But the situation seems rather strange when one wants to find an equiconsistent theory with $ZFC$ which has end points in both up and down direction because the existence of two contradictory objects like $\emptyset$ and $G$ seems ontologically incompatible in a particular "$ZFC$-like" world. So the question is:

Question (1): Is there an $\mathcal{L}=\lbrace \in\rbrace$-theory $T$ such that the following conditions hold:

$(1)~Con(ZFC)\Longleftrightarrow Con(T)$

$(2)~T\models \exists !x~\forall y~~(y\in x)$

$(3)~T\models \exists !x~\forall y~~\neg (y\in x)$

Remark (1): Quine's new foundation axiomatic system ($NF$) is not an answer because its equiconsistency with $ZFC$ is still unknown.

Even one can define two dual sets from empty and global sets. The set which does not belong to any other set ($\emptyset^{\star}$) and the set which belongs to any set ($G^{\star}$).Now one can restate the question (1) as follows:

Question (2): Is there an $\mathcal{L}=\lbrace \in\rbrace$-theory $T$ such that the following conditions hold:

$(1)~Con(ZFC)\Longleftrightarrow Con(T)$

$(2)~T\models \exists !x~\forall y~~(x\in y)$

$(3)~T\models \exists !x~\forall y~~\neg (x\in y)$

Even it is interesting to have an equiconsistent theory which has no end points in both up and down directions.So:

Question (3): Is there an $\mathcal{L}=\lbrace \in\rbrace$-theory $T$ such that the following conditions hold:

$(1)~Con(ZFC)\Longleftrightarrow Con(T)$

$(2)~T\models \neg (\exists x~\forall y~~(y\in x))$

$(3)~T\models \neg (\exists x~\forall y~~\neg (y\in x))$