elementary embeddings, $V$ vs. $L$

Neither option is possible.

The key point is that "being $L$" is first-order expressible: since $L\models$ $\mathsf{V=L}$, if there is an elementary embedding either way between $V$ and $L$ we must have $V\models\mathsf{V=L}$ and so $V=L$. But by the Kunen inconsistency, there is no elementary embedding from $V$ to itself.

A couple comments:

• Re: $V$, $L$, and $\mathsf{V=L}$ above, this is not a typo: I reserve the "mathsf" font for axioms and theories, and normal italicized Roman letters for objects.

• Invoking Kunen here is overkill: you can prove more elementarily from $\mathsf{ZFC+V=L}$ that there is no (nontrivial) elementary embedding $L\rightarrow L$ (while on the other hand, the existence of an elementary embedding $L\rightarrow L$ follows from a measurable cardinal or indeed much less). However, I think bringing Kunen into the picture is helpful here since it presents a more unavoidable theme.

• The Kunen inconsistency requires choice (to the best of our current knowledge), and so one might hope to avoid the argument above if we merely assume $\mathsf{ZF}$. But remember that in fact $\mathsf{ZF}\vdash (L\models \mathsf{AC+V=L})$, so this doesn't help.