A finite algebra $(X,*,1)$ is a reduced Laver-like algebra if it satisfies the identities $x*(y*z)=(x*y)*(x*z)$ and if there is a surjective function $\mathrm{crit}:X\rightarrow n+1$ where
$\mathrm{crit}(x)=n$ if and only if $x=1$
$\mathrm{crit}(x*y)=\mathrm{crit}(y)$ whenever $\mathrm{crit}(y)<\mathrm{crit}(x)$, and
$\mathrm{crit}(y)<\mathrm{crit}(x*y)$ whenever $\mathrm{crit}(x)\leq \mathrm{crit}(y)<n$.
The number $n+1$ and the function $\mathrm{crit}$ are uniquely determined by $(X,*,1)$. If $\mathrm{crit}:X\rightarrow n+1$, then we say that $X$ has $n+1$ many critical points. If $x\in X$, then define a mapping $x^{\sharp}:n+1\rightarrow n+1$ by letting $x^{\sharp}(\mathrm{crit}(y))= \mathrm{crit}(x*y)$.
Suppose that $\mathcal{X}=((X_{n})_{n\in\omega},(\phi_{n,m})_{n\geq m})$ is an inverse system of finite Laver-like algebras where each $X_{n}$ has $n+1$ many critical points, each transitional mapping $\phi_{n,m}$ is surjective, and where if $c\in X_{n}$ and $\mathrm{crit}(c)=n-1$, then $\phi_{n,n-1}(x)=\phi_{n,n-1}(y)$ if and only if $c*x=c*y$.
Suppose furthermore that if $x,y\in X_{n}$, there is some $N\geq n$ along with $x',y'\in X_{N}$ where $x'*y'\neq 1$ and where $\phi_{N,n}(x')=x,\phi_{N,n}(y')=y$.
Intuitively, the inverse system $((X_{n})_{n\in\omega},(\phi_{n,m})_{n\geq m})$ is an algebraization of the inverse system $(\mathcal{E}_{\lambda}/\equiv^{\alpha_{n}})_{n\in\omega}$ of quotient algebras of rank-into-rank embeddings where $(\alpha_{n})_{n\in\omega}$ is a cofinal increasing collection of limit ordinals in $\lambda$.
Now, if $j:V_{\lambda}\rightarrow V_{\lambda}$ is an elementary embedding, then $(j*j)(\alpha)\leq j(\alpha)$ for each $\alpha<\lambda$. Strangely enough, there exists a Laver-like algebra $X$ with $|X|=11$ with $6$ critical points and $x\in X$ where $(x*x)^{\sharp}(\alpha)<x^{\sharp}(\alpha)$.
Define $\mathbf{EE}(\mathcal{X})$ to be the subalgebra of $\varprojlim_{n\in\omega}X_{n}$ consisting of all threads $(x_{n})_{n\in\omega}\in\varprojlim_{n\in\omega}X_{n}$ where for all $N$ there exists an $M\geq N$ and a thread $(y_{n})_{n\in\omega}\in\varprojlim_{n\in\omega}X_{n}$ where $y_{N}=1$ but where $x_{M}*y_{M}\neq 1$. The set $\mathbf{EE}(\mathcal{X})$ is an algebraization of the collection of all elementary embeddings in $\mathcal{E}_{\lambda}$.
If $x=(x_{n})_{n\in\omega}\in\mathbf{EE}(\mathcal{X})$, then define a function $x^{\sharp}:\omega\rightarrow\omega$ by letting $x^{\sharp}(n)=m$ if whenever $N>m$, we have $x_{N}^{\sharp}(n)=m$.
Is it possible that $x^{\sharp}(n)<(x*x)^{\sharp}(n)$ for infinitely many $n$? Is it possible that for each $m$ there are infinitely many $n$ with $x^{\sharp}(n)<(x^{[2]})^{\sharp}(n)<\dots<(x^{[m]})^{\sharp}(n)$?
The motivation for this question is that I want to know how badly can $\mathbf{EE}(\mathcal{X})$ fail to resemble any possible $\mathcal{E}_{\lambda}.$
If for all $n$ there is an $n$-huge cardinal, then for all $x\in\mathbf{EE}(\mathcal{X})\setminus\{(1)_{n\in\omega}\}$ and each term $t$, we have $$(x*x)^{\sharp}(\mathrm{crit}(t(x)))\leq x^{\sharp}(\mathrm{crit}(t(x))).$$ Furthermore, for each $x\in X$, we have $(x*x)*x^{[n]}=x*x^{[n]}$ whenever $n>1$, so there are infinitely many $n$ with $x^{\sharp}(n)=(x*x)^{\sharp}(n)$.
