I came across the following toy problem and was curious if there was a simple solution or counterexample. Suppose you have a distribution $p$ on $m$ random variables $X_1, \ldots, X_m$, each with finite alphabet of size $a$, satisfying
(i) for some $n<m$, $X_{n+1}, \ldots, X_m$ are i.i.d. with the uniform distribution (as well as independent from $X_1,\ldots, X_n$), and
(ii) $X_1,\ldots,X_n$ are $k$-wise independent, and again each marginal is distributed uniformly.
Consider the distribution formed by first choosing $(X_1,\ldots,X_m) \sim p$ and then randomly permuting the coordinates. Is the resulting distribution close (in total variation) to the uniform distribution for some choice of parameters $m,n,k,a$?
I've convinced myself that in the best case: for $k=n-1$ and $m =\operatorname{poly}(n)$, one can expect that the statistical distance between the resulting distribution and the uniform distribution is negligible in $n$, even with the alphabet size exponential in $n$. To see this, consider the case where the $X_i$ is drawn from $\mathbb{F}_2^n$ and $X_n = \sum_{i=1}^{n-1} X_i$.
