Suppose that $\beta$ is a $2n$-strand braid with plat closure $L$. We can multiply $\beta$ on either side by a member of the Hilden subgroup to get a new braid whose plat closure is still $L$. Or we could replace $\beta$ with $\beta\sigma_{2n-1} \in B_{2n+2}$ to get another braid whose plat closure is $L$. (Here $\sigma_{2n-1}$ is one of the usual Artin generators.)

In "On the Stable Equivalence of Plat Representations of Knots and Links", Birman proves that these moves are 'sufficient'.

Theorem:Let $K$ and $K'$ be knots. Let $\alpha$ and $\beta$ be braids whose plat closures are isotopic to $K$ and $K'$ respectively. Then $K$ is isotopic to $K'$ if and only if there are stabilizations $\alpha'$ and $\beta'$ of $\alpha$ and $\beta$ which lie in the same Hilden double coset.

(The theorem is still true for links with one stabilization operation for each component of the closure.)

Montesinos exhibited two minimal index braids whose plat closures are isotopic but which do not lie in the same Hilden double coset. So stabilization is necessary.

Now suppose that $\beta$ and $\beta'$ are plat representations of the $k$-component unlink. Is anything known about whether stabilization is necessary?