For this question, suppose that there exists a rank-into-rank cardinal. Let $\mathcal{E}_{\lambda}$ be the set of all elementary embeddings $j:V_{\lambda}\rightarrow V_{\lambda}$. Give $\mathcal{E}_{\lambda}$ the operation $*$ defined by $j*k=\bigcup_{\alpha<\lambda}j(k|_{V_{\alpha}}))$. Define $j_{[n]}$ for all $n$ and $j\in\mathcal{E}_{\lambda}$ inductively by letting $j_{[1]}=j,j_{[n+1]}=j_{[n]}*j$. If $\gamma<\lambda$ is a limit ordinal, then let $\equiv^{\gamma}$ be the congruence on $\mathcal{E}_{\lambda}$ defined by $j\equiv^{\gamma}k$ iff $j(x)\cap V_{\gamma}=k(x)\cap V_{\gamma}$ for each $x\in V_{\gamma}$. Let $A_{n}$ denote the $n$-th classical Laver table. Recall that $A_{n}=(\{1,...,2^{n}\},*)$ where $*$ is defined by
$x*1=x+1\mod 2^{n}$, and
$x*(y*z)=(x*y)*(x*z)$.
Give $\varprojlim A_{n}$ the Borel probability measure $\mu$ where for $i\in\{1,...,2^{N}\}$ we have $\mu(\{(x_{n})_{n}|x_{N}=i\})=(\frac{1}{2})^{N}$.
Let $j\in\mathcal{E}_{\lambda}$. Let $E\subseteq\varprojlim A_{n}$ be the set of all $(x_{n})_{n}$ such that there is some $k\in\mathcal{E}_{\lambda}$ where $k\equiv^{\kappa_{n}}j_{[x_{n}]}$ for all $n\in\omega$ where $\kappa_{n}$ is the $n$-th element in $S=\{\text{crit}(l)|l\in\langle j\rangle\}$ (for example, $crit(j)$ is the $0$-th element in $S$).
What is the measure $\mu(E)$? Is $\mu(E)=0$?
$\mathbf{Note:}$ As with many questions about the algebras of elementary embeddings, one can formulate this question in terms of algebra and topology without any mention of large cardinals.
Let $U_{m}$ be the set of all $(x_{n})_{n\in\omega}\in\varprojlim A_{n}$ such that there is some $n$ where $x_{n}*_{n}2^{m}<2^{n}$. Then $U_{m}$ is open in $\varprojlim A_{n}$ and $E=\bigcap_{m}U_{m}$ which is a $G_{\delta}$-set and if a rank-into-rank cardinal exists, then $E$ is dense in $\varprojlim A_{n}$.
