For permutation $\pi:[n]\mapsto[n]$ let $L(\pi)$ denote the length of longest increasing sub-sequence of $\pi$, and ${S}_{n,m}$ be the set of all permutations with longest increasing sub-sequence of $m$ \begin{align} &L(\pi):= \max\{k: \pi(i_1)<\dots<\pi(i_k)\; \text{for some}\; 1\le i_1<\dots<i_k\le n\}\\ &{S}_{n,m}:=\{\pi:[n]\mapsto[n]\; \vert\; L(\pi)=m\} \end{align} Is there a way to uniformly sample permutations from this set $\pi\sim \mathrm{Unif}({S}_{n,m})$?
UPDATE thanks to Phicar's comment, I came across Robinson–Schensted correspondence, and came up with the following randomized procedure, assuming that $m \ge C$ for some absolute constant $C$: Draw $\mathbf{Z}$ uniformly from $m$-dimensional unit cube and sort its elements into $\mathbf{X}$. Draw the rest of elements from open interval $(X_1,X_m)$, and define permutation $\pi:[n]\mapsto [n]$ as the rank of elements in $X$: \begin{align} &\mathbf{Z}\sim\mathrm{Unif}([0,1]^m), && (X_i)_{i\le m}\gets \mathrm{sort}(Z_i)_{i\le m}\\ &X_{m+1},\dots,X_n \sim \mathrm{Unif}(X_1,X_m), &&\pi(i) \gets \#\{j\in[i]\colon X_j\le X_i)\} \end{align}
Explanation: the order of points $X_i$s uniquely determines the permutation, at step one we start with Young tableau that is horizontal row with $m$ elements in the row. Since the elements are generated strictly between the minimum $X_1$ and maximum $X_m$, the length of $\lambda_1$ will never increase during the procedure. Therefore, this corresponds to a random-walk on Young lattice.
However, does this impose a uniform distribution on $S_{n,m}$?