I forgot to mention originally: this was motivated by this old MSE question.
It's not hard to show in $\mathsf{ZFC}$ that there are nontrivial elementary embeddings from $(Ord; \in)$ to itself - or rather, it's not hard to write down specific formulas which $\mathsf{ZFC}$ proves define such maps. When we add common algebraic operations, things get messier but the picture seems to stay the same - e.g. the result continues to hold for $Ord$ equipped with ordinal addition, multiplication, and exponentiation in addition to $\in$. It seems like any "reasonably simple" expansion of $Ord$ has $\mathsf{ZFC}$-provable nontrivial elementary self-embeddings.
I'm interested in a particular precisiation of this intuition. Specifically, the following seems very close to the Kunen inconsistency but I can't quite get it there:
Is it consistent that there is a nontrivial second-order elementary self-embedding $f:(Ord;\in)\rightarrow (Ord;\in)$?
Note that the second-order quantifiers in this case range over subclasses of $Ord$ or its finite Cartesian powers, not just subsets (= bounded subclasses). So we really have to ask this within an appropriate class theory. Given that that may be a bit annoying to consider, here's a "set-ified" version of that question:
Is it consistent with $\mathsf{ZFC}$ that there is an inaccessible $\kappa$ and a second-order nontrivial elementary embedding $f:(\kappa;\in)\rightarrow(\kappa;\in)$?
(Here correspondingly the second-order quantifiers range over the $\mathcal{P}(\kappa^{n})$s for $n\in\omega$ as appropriate, so everything is nicely captured by $V_{\kappa+2}$: $f$ itself, or an equivalent coding thereof, lives in $V_{\kappa+1}$, the evaluation of each second-order formula over $V_{\kappa}$ is determined by $V_{\kappa+1}$, and $V_{\kappa+2}$ can express "is a second-order elementary embedding" directly.)
The difficulty I run into in getting a negative answer here by invoking the Kunen inconsistency is that even using second-order logic there doesn't seem to be a way to code $V$ into $Ord$ (or $V_\kappa$ into $\kappa$). Of course if $V$ is "sufficiently canonical" we can do this, e.g. the existence of such an $f$ (or such a $\kappa$ and $f$) clearly contradicts $\mathsf{V=L}$, but I don't see an argument for the general case.