Suppose two sequences give the same list. Sort them as

$a_1\le\cdots\le a_n,\qquad b_1\le\cdots\le b_n.$

If they differ, let $k$ be the first differing index and, after swapping the sequences if necessary, assume $a_k<b_k$.

From each list, remove the averages of all nonempty subsets of indices $1,\ldots,k-1$, one copy per subset. These removed multisets are equal, so the remaining multisets are equal too. Each remaining multiset consists exactly of the averages of subsets meeting indices $k,\ldots,n$.

Let $m$ be their smallest entry. In the $b$-sequence, choose a subset with average $m$ that meets $k,\ldots,n$ and has the fewest elements. Since the singleton $\{k\}$ is eligible, $m\le b_k$.

This subset contains exactly one index from $k,\ldots,n$. Indeed, if it contained at least two, remove one of their entries, say $y$. Since $y\ge b_k\ge m$, removing it cannot increase the average, and the subset still meets $k,\ldots,n$. A smaller average contradicts the definition of $m$; an equal average contradicts the chosen minimum size.

Replace its unique entry from this suffix by $a_k$, and keep the other indices, all of which are below $k$. This gives a subset of the $a$-sequence meeting the suffix. Its average is strictly smaller than $m$, because the replaced entry was at least $b_k>a_k$. This contradicts the minimum of the remaining multiset.

Thus the sorted sequences are equal, proving uniqueness up to permutation.