An algebraic structure $(X,*)$ is said to be self-distributive if it satisfies the identity $x*(y*z)=(x*y)*(x*z)$.

Suppose that $X$ is a self-distributive algebra. Then the positive braid monoid $B_{n}^{+}$ acts on $X^{n}$ by letting $(x_{1},...,x_{n})\cdot\sigma_{i}=(x_{1},...,x_{i-1},x_{i}*x_{i+1},x_{i},x_{i+2},...,x_{n})$ and this action is called the Hurwitz action.

Recall that an algebraic structure $(X,*)$ is left-cancellative if $x*y=x*z$ implies $y=z$. If $X$ is left-cancellative, then this action extends to a partial action of the braid group $B_{n}$ on $X^{n}$ by defining $(x_{1},...,x_{n})\cdot bc^{-1}=(y_{1},...,y_{n})$ precisely when $(x_{1},...,x_{n})\cdot b=(y_{1},...,y_{n})\cdot c$, and this partial action is still called the Hurwitz action.

Let $\mathcal{E}_{\lambda}^{+}$ be the set of all elementary embeddings $j:V_{\lambda}\rightarrow V_{\lambda}$. Then $\mathcal{E}_{\lambda}^{+}$ can be endowed with a self-distributive operation $*$ defined by letting $j*k=\bigcup_{\alpha<\lambda}j(k|_{V_{\alpha}})$. The operation $*$ is not associative nor commutative though. The cardinal $\lambda$ is the limit of countably many very large cardinals. For more information about the algebraic structure $(\mathcal{E}_{\lambda}^{+},*)$, please see chapter 11 from the Handbook of Set Theory. For information about both braids, the Hurwitz action, and elementary embeddings, please see chapters 1,3,12 from the book Braids and Self-Distributivity by Patrich Dehornoy.

Suppose that $j_{1},...,j_{n}\in\mathcal{E}_{\lambda}^{+}$ and $\gamma<\lambda$. Recall that $\mathrm{crit}(j)$ denotes the smallest ordinal $\alpha$ where $j(\alpha)>\alpha$. Then are there only finitely many braids $b\in B_{n}$ such that if $(j_{1},...,j_{n})\cdot b=(k_{1},...,k_{n})$, then $\mathrm{crit}(k_{1})<\gamma,...,\mathrm{crit}(k_{n})<\gamma$? If we restrict ourselves to the positive braids, then the answer is a yes by my answer here.