The answer is yes, the existence of such embeddings is equiconsistent over ZFC with a measurable cardinal.

If $\kappa$ is measurable, then there is a fully elementary embedding $j:V\to M$ into a transitive class $M$ with critical point $\kappa$. By composing this map with the inclusion $M\subset V$, we may view $j:V\to V$. The observation to make is that this remains $\Delta_0$-elementary, since every transitive class is $\Delta_0$-elementary in the universe $M\prec_{\Sigma_0}V$, and so we may view $j:V\to V$ as a composition of the original elementary embedding with this $\Delta_0$-elementary inclusion.

In fact, the embedding $j:V\to V$ is $\Delta_1$-elementary, since $M$ is actually $\Delta_1$-elementary in $V$, as $\Sigma_1$ assertions go up and $\Pi_1$ assertions go down. Since $M$ and $V$ are elementarily equivalent, they agree on the equivalence or non-equivalence of any two formulas.

(Note that there is a subtle issue about what exactly one means by $\Delta_1$-elementary, since there is no such thing as a $\Delta_1$-formula, unlike $\Delta_0$. For example, in the argument, it was important that $M$ and $V$ agreed on the equivalence of the $\Sigma_1$ nad $\Pi_1$ formulas in question. It can happen, however, that two models of ZFC do not necessarily agree like this, and there is a difference between $\Delta_1$ and provably-$\Delta_1$.)

Conversely, let us show that measurable cardinals are required. If $j:V\to V$ is a nontrivial $\Delta_0$-elementary embedding, then I claim, first, that there is a critical point, that is, a first ordinal that is moved. Notice that $j$ must take ordinals to ordinals. Assume toward contradiction that $j$ fixes all ordinals. Let $a$ be an $\in$-minimal set with $j(a)\neq a$. So $a\subset j(a)$ since $j(x)=x$ for all $x\in a$. By the axiom of choice, we may well-order $a$ with a relation $\lhd$ in some order-type $\gamma=|a|$. It follows that $j(\lhd)$ well-orders $j(a)$ in order-type $j(\gamma)=\gamma$. Furthermore, the $\beta^{th}$ element of $a$ gets mapped to the $j(\beta)=\beta^{th}$ element of $j(a)$ with respect to $j(\lhd)$. So $j$ is onto $j(a)$, and therefore $j(a)=j"a=a$, a contradiction. So we have shown that there must be a critical point $\kappa$, the least ordinal with $\kappa<j(\kappa)$. Secondly, now, let $\mu$ be the set of $X\subseteq\kappa$ with $\kappa\in j(X)$. It follows now that $\mu$ is a $\kappa$-complete nonprincipal ultrafilter on $\kappa$. This is easily seen to be a nonprincipal filter, and it is $\kappa$-complete, since if $X_\alpha\in \mu$ for all $\alpha<\gamma$, some $\gamma<\kappa$, then $j(\langle X_\alpha\mid\alpha<\gamma\rangle=\langle j(X_\alpha)\mid\alpha<\gamma\rangle$, and $\kappa$ is in every element of the right hand side and hence in the intersection of the right-hand side. So $\cap_\alpha X_\alpha\in\mu$.

Lastly, let me point out that if one weakens much more, below $\Delta_0$, then one gets the embeddings in ZFC. This idea grew out of my work in:

*Hamkins, Joel David*, Every countable model of set theory embeds into its own constructible universe, J. Math. Log. 13, No. 2, Article ID 1350006, 27 p. (2013). DOI, ZBL1326.03046,
blog.

In that article, I consider the $\in$-embeddings, which are embeddings $j$ for which merely $$x\in y\iff j(x)\in j(y).$$ This is a weakening of $\Delta_0$-elementarity. These are precisely the self-embeddings of the structure $\langle V,\in\rangle$ in the model-theorists sense of the term for a relational structure.

**Theorem.** ZFC proves that the map defined by the following recursion is an $\in$-embedding $j:V\to V$.
$$j(x)=\{j(y)\mid y\in x\}\cup\{\{\emptyset,x\}\}$$

See theorem 19 and lemma 18 in the paper. Basically, it is clear that $y\in x\to j(y)\in j(x)$ by definition. And conversely, $j$ is injective, since we can recover $x$ from the last part of $j(x)$, and if $j(y)\in j(x)$, then it must have come from $y\in x$, since $j(y)\neq\{\emptyset,x\}$ as $j(z)\neq\emptyset$ and $\emptyset\notin j(z)$ for any $z$.

Let me also mention my question, Can there be an embedding $j:V\to L$?, which asks whether one can have in GBC an $\in$-embedding from $V$ to $L$, in cases where $V\neq L$. The full question remains open, although we now know some things about partial answers, for work-in-progress.