Let $\mathcal{E}_{\lambda}$ be the set of all elementary embeddings $j:V_{\lambda}\rightarrow V_{\lambda}$. If $A\subseteq V_{\lambda}$, then let $j^{+}(A)=\bigcup_{\alpha<\lambda}j(A\cap V_{\alpha})$, and let $\mathcal{E}_{\lambda}[A]=\{j\in\mathcal{E}_{\lambda}\mid j^{+}(A)=A\}.$

We say that a non-trivial subalgebra $X\subseteq\mathcal{E}_{\lambda}$ has the simplicity property if whenever $\gamma<\lambda$ and $\gamma$ is a limit ordinal, and $\simeq$ is a congruence on $X$ such that $j\equiv^{\gamma}k\Rightarrow j\simeq k$ for all $j,k\in X$, then there is some $\ell\in\mathcal{E}_{\lambda}[A]$ where $j\simeq k$ if and only if $j\equiv^{\mathrm{crit}(\ell)}k$ for all $j,k\in X$.

Suppose that $V_{\lambda}\models V=HOD$. Then does every subalgebra $X\subseteq\mathcal{E}_{\lambda}$ satisfy the simplicity property? Does every finitely generated subalgebra $X\subseteq\mathcal{E}_{\lambda}$ satisfy the simplicity property?

Let $A$ be a well-ordering (or linear ordering) of $V_{\lambda}$ and suppose that $X\subseteq\mathcal{E}_{\lambda}[A]$ is a subalgebra (or a finitely generated subalgebra). Then does $X$ necessarily satisfy the simplicity property?