Let $B_{\infty}$ denote the infinite strand braid group. Let $\mathrm{sh}:B_{\infty}\rightarrow B_{\infty}$ be the mapping where $\mathrm{sh}(\sigma_{i})=\sigma_{i+1}$ whenever $i\geq 1$. Then $B_{\infty}$ can be endowed with an operation $*$ known as shifted conjugacy where $*$ is defined by $$x*y=x\cdot\mathrm{sh}(y)\sigma_{1}\mathrm{sh}(x)^{-1}.$$ Then $*$ satisfies the self-distributivity identity $$x*(y*z)=(x*y)*(x*z).$$
A rack is an algebra $(X,*,*^{-1})$ that satisfies the following identities:
$x*(y*z)=(x*y)*(x*z)$ and
$x*(x*^{-1}y)=x*^{-1}(x*y)=y$.
A function $T:X^{2}\rightarrow X^{2}$ is said to satisfy the Yang-Baxter equation if $$(1_{X}\times T)\circ(T\times 1_{X})\circ(1_{X}\times T)=(T\times 1_{X})\circ(1_{X}\times T)\circ(T\times 1_{X}).$$
For example, if $(X,*,*^{-1})$ is a rack and $T(x,y)=(x*y,x)$, then $T$ always satisfies the Yang-Baxter equation.
If $T:X^{2}\rightarrow X^{2}$ satisfies the Yang-Baxter equation, then define an action of $B_{n}$ on $X^{\mathbb{N}}$ by letting $$(x_{1},\dots,x_{n},)\cdot \sigma_{i} =(x_{1},\dots,x_{i-1},T(x_{i},x_{i+1}),x_{i+2},\dots,x_{n}).$$
Suppose that $X$ is a finite set and $T:X^{2}\rightarrow X^{2}$ is a bijective function that satisfies the Yang-Baxter equations. Then is the function $$(x_{1},...,x_{k},t,a_{1},...,a_{n})\mapsto(x_{1},...,x_{k})\cdot t(a_{1},...,a_{n})$$ where $a_{1},...,a_{n}\in B_{\infty},x_{1},...,x_{k}\in X$, and $t$ is a term in the language of self-distributive algebras necessarily computable in polynomial time? What about the special case when $T(x,y)=(x*y,x)$ for some rack $(X,*,*^{-1})$ or when we restrict our case to when $t$ is a term in only one variable?
