= Solution
The <braid group> $B_n$ has generators $\sigma_1,\ldots,\sigma_{n-1}$ and relations
$$
\sigma_i\sigma_j=\sigma_j\sigma_i\ (|i-j|>1),\qquad\sigma_i\sigma_{i+1}\sigma_i=\sigma_{i+1}\sigma_i\sigma_{i+1}.
$$
In the <strict monoidal category> $\mathcal C$, put $X^{\otimes0}=I$ and send
$$
\boxed{n\longmapsto X^{\otimes n},\qquad\sigma_i\longmapsto1_X^{\otimes(i-1)}\otimes y\otimes1_X^{\otimes(n-i-1)}.}
$$
The images are invertible because the <Yang–Baxter operator> $y$ is invertible. Distant images commute by the tensor interchange law, and adjacent images satisfy the <braid group relations> by the <Yang–Baxter operator> equation tensored with the remaining identities. Thus these assignments define <group homomorphisms> $B_n\to\operatorname{Aut}(X^{\otimes n})$, and hence a <functor> $\mathbf B\to\mathcal C$.
Juxtaposition of braids puts generators into the corresponding two blocks of tensor factors, so their images are the tensor products of their images. Strictness of $\mathcal C$ gives $X^{\otimes(m+n)}=X^{\otimes m}\otimes X^{\otimes n}$ and $X^{\otimes0}=I$ exactly. \b[The resulting functor is strict monoidal and sends $1$ to $X$.]
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