Suppose that $j_{1},\dots,j_{n}:V_{\lambda+1}\rightarrow V_{\lambda+1}$ are elementary embeddings. Then does there necessarily exist a linear ordering $A$ of $V_{\lambda}$ such that $j_{1}(A)=\dots=j_{n}(A)=A$? The result holds if $V_{\lambda}\models(V=HOD)$ which is consistent with the existence of an elementary embedding $j:V_{\lambda+1}\rightarrow V_{\lambda+1}$, but does the result hold in general without any additional hypotheses? I am also interested in the version of this question when $j_{1},\dots,j_{n}$ are only elementary embeddings from $V_{\lambda}$ to $V_{\lambda}$.

The case when $n=1$ is easy since if $j:V_{\lambda+1}\rightarrow V_{\lambda+1}$ is an elementary embedding and if $B$ is a linear ordering on $V_{\mathrm{crit}(j)}$, then $A=\bigcup_{n}j^{n}(B)$ is a linear ordering on $V_{\lambda}$ with $j(A)=A$. I am therefore looking at whether the result holds in general instead of particular instances of the result.