[EDIT] the prior question (see the second section below) was trivially false, however the intention is to *arrange* a possible world of such universes, in other words the question is about if it is possible to have a proper class $\mathcal {G}$ of Grothenderick universes that contains $V$ in them, such that every object that is an element of some class (i.e. a set) then it is in $\bigcup(\mathcal{G})$, here $\mathcal{G}$ is a primitive constant symbol. So the formulation would be:


if $\varphi(a\_1,..,a\_n)$ is a formula in $L(\in,=)$, in which all and only $a\_1,...,a\_n$ occur free, then:


$\forall W \big{(} W \in \mathcal {G} \to \\\forall a\_1,..,a\_n \in V [\text{if } \langle W,\in \rangle \models \varphi(a\_1,..,a\_n) \text{ then } \langle V,\in \rangle \models \varphi(a\_1,..,a\_n)] \big{)}$




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The older question was


Add a constant primitive symbol $V$ to the language of $ZF$, standing for some "sub-world" of sets.Now axiomatize that $V$ is a Grothendieck universe. Add Tarki's axiom of every object is an element of some Grothendieck universe. Now is it possible to arrange that all Grothendieck universes containing $V$ do not add additional theorems about elements of $V$ in the pure set theoretic language (i.e. in $L(\in,=)$)?


Formally:
if $\varphi(a\_1,..,a\_n)$ is a formula in $L(\in,=)$, in which all and only $a\_1,...,a\_n$ occur free, then:


$\forall W \big{(}W \text { is a Grothendieck Universe } \wedge V \in W \to \\\forall a\_1,..,a\_n \in V [ \langle W,\in \rangle \models \varphi(a\_1,..,a\_n) \implies \\\langle V,\in \rangle \models \varphi(a\_1,..,a\_n)] \big{)}$


The second question can we upgrade that to be from free variables in $W$? i.e. can we have the following?


if $\varphi(a\_1,..,a\_n)$ is a formula in $L(\in,=)$, in which all and only $a\_1,...,a\_n$ occur free, then:


$\forall W \big{(}W \text { is a Grothendieck Universe } \wedge V \in W \to \\\forall a\_1,..,a\_n \in W [ \langle W,\in \rangle \models \varphi(a\_1,..,a\_n) \implies \\\langle V,\in \rangle \models \varphi(a\_1,..,a\_n)] \big{)}$



Can there be elementary embedding between a universe and a universe inside it?
No, even the "weak" version (indeed, a weakening of that) is impossible. 


Consider the sentence *(no parameters needed!)* $$\mbox{$\varphi\equiv$ "There is a largest inaccessible cardinal."}$$ If there is a proper class of inaccessible cardinals, then for proper class many inaccessible $\kappa$ we have $V\_\kappa\models\varphi$ and for proper class many inaccessible $\kappa$ we have $V\_\kappa\models\neg\varphi$. So the "theory of the inaccessibles" can *never* stabilize.


*Remember that the Grothendieck universes are exactly the $V\_\kappa$s for $\kappa$ inaccessible, so we're really just talking about the behavior of inaccessible cardinals. In particular, "Every set is contained in a Grothendieck universe" is equivalent to "There is a proper class of inaccessible cardinals."*


*EDIT*: A better counterexample sentence might be $$\psi\equiv\mbox{"The class of inaccessibles has ordertype $\alpha\cdot 2$ for some ordinal $\alpha$."}$$


*(In particular, note that $\psi$ implies that there are only set-many inaccessibles.)* The reason this is preferable is that it works in a broader context: if we weaken our background theory to allow models with a proper class of inaccessibles of ordertype $\omega$ (that is, kill replacement), the $\varphi$ above doesn't work, but this $\psi$ still does. Indeed, this $\psi$ kills *every* version of the original question that makes sense, as far as I can tell.




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Note that your title talks about *elementary embeddings*, but the above argument kills even the possibility of mere elementary *equivalence* between all sufficiently-large universes. Also, note that we could replace "inaccessible" with any definable large cardinal notion and get the same phenomenon. However, the word "definable" there is important. For example, if $0^\sharp$ exists then we have $L\_\alpha\prec L\_\beta$ whenever $\alpha<\beta$ are uncountable cardinals in the sense of $V$.




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Re: your edit, the answer is now **yes**. Essentially - again, shifting from *universes* to *cardinals* (I really think this makes everything much easier to think about) - what you're basically asking is whether we can have a cofinal family of indiscernible inaccessible cardinals. As per my comment above, one way to get such a model is the following: suppose $M\models$ "There is a measurable cardinal," and consider the structure $N=(L^M; \in, UncCard^M)$ where $UncCard^M$ is the class of uncountable cardinals in $M$. We can now take $W\_0$ *(what you call "$V$," but I think that leads to overloading)* to be $(V\_\kappa)^M$ for any element $\kappa$ of $UncCard^M$ in this context (fine, after "throwing out" all elements of $UncCard^M$ which are $<\kappa$).


EDIT: And as Monroe Eskew observes we can do massively better than a measurable: if $\kappa$ is Mahlo, then $V\_\kappa$ gives a model of the type you want.


*And I believe even* that *is overkill: I suspect that, if we take a countable model $M$ of ZFC+"There is a proper class of inaccessbile cardinals" then the ultrapower of $M$ along any nonprincipal ultrafilter on $\omega$ has a cofinal indiscernible family of inaccessibles. However, I haven't checked the details, so I could be wrong. Also, note that even if this works it gives an illfounded model, whereas Monroe's observation gives a wellfounded model, admittedly from a significantly stronger assumption.*


