New toy conjecture on randomly sampled permuted Reed–Muller puzzles

Let $\mathbb{F}$ be a finite field with $|\mathbb{F}|=q\approx\lambda^2$. Let $p_1,\ldots,p_m$ be uniformly random polynomials of degree at most $\lambda$ in $\mathbb{F}[X]$, where $m=\lambda^{100}$, and let $q_1,\ldots,q_m$ be uniformly random functions from $\mathbb{F}$ to $\mathbb{F}$. Let $\pi\in S_{\mathbb{F}\times\mathbb{F}}$ be a uniformly random permutation, and let $\Omega^{(1)},\ldots,\Omega^{(m)}\subset\mathbb{F}$ be independent uniformly random sets of size $100\cdot\lambda$. Define

$$S_i=\left\{\pi(x,p_i(x)):x\in\Omega^{(i)}\right\},\qquad T_i=\left\{\pi(x,q_i(x)):x\in\Omega^{(i)}\right\}.$$

New toy conjecture. The distributions $(S_1,\ldots,S_m)$ and $(T_1,\ldots,T_m)$ are computationally indistinguishable.

This replacement is designed to resist the efficient rank-based attack that refutes the original toy conjecture, because the evaluation points are independently resampled as random subsets for each $i$. The source presents it as a proposed replacement; no resolution is given here.