Are there meaningful elementary embeddings of transitive models of set theory without a critical point? Is it possible to have $j:M\to N$ elementary, where M,N are transitive set models of ZFC, with $j\mid\operatorname{Ord}^M = \operatorname{id} \mid \operatorname{Ord}^M$, but there is $x\in M$ with $j(x)\neq x$ ? I know that if $j(X)\subseteq M$ for any set $X\in M$ of ordinals (and $\operatorname{Ord}^M=\operatorname{Ord}^N$ would be sufficient to prove this), then $j$ has a critical point, so we would need to have $\operatorname{Ord}^M \subsetneq \operatorname{Ord}^N$ , i.e. $j$ is inclusion on ordinals, but not on sets. In what context would such $j$ arise? Maybe one could take an elementary substructure of a ZFC model and shift it around a little, while preserving the elementarity? I'm not sure.
 A: I found out that the answer somewhat depressingly is no: Consider $Y = \bigcup j'' M$. This $Y$ is transitive, because for any $y_1\in y_0\in j(x)$, it is true that $y_1\in j(\operatorname{trcl}(x))$.
And $Y$ is an elementary $\in$-substrucure of $N$: 
If for some $\varphi$, $N\models \varphi(z,y)$, where $y\in j(x)$, then $N\models \exists z' \ \psi(z',j(x))$, where 
$$\psi(z',v)=  \forall y'\in v\, (\exists z''  \varphi(z'',y') \rightarrow \exists z''\in z' \varphi(z'',y'))$$
So $M\models \exists z'\ \psi(z',x)$. Let $x'\in M$ be such. Then $N\models \psi(j(x'),j(x))$, and so there is $z'\in j(x')$ with $N\models \varphi(z',y)$.
Then $j:M\to \bigcup j'' M$ is an elementary embedding of transitive ZFC models with $\alpha:= \operatorname{On} \cap M = \operatorname{On}\cap \bigcup j'' M$ and $j\mid\alpha = \operatorname{id} \mid \alpha$, and therefore $\forall x\in M \ j(x)=x$ by the usual argument: $j$ acts like the identity on sets of ordinals and thus by the axiom of choice in $M$, also on sets.
