Here is a more general (and possibly more clearly stated) combinatorial version of this question, than this question. Let $u$ be a sequence of 0 and 1 of length $n$. Let $p,q,r$ be three natural numbers, $p+q+r=n$, and $w$ be a sequence of $a, b, c$ started with $p$ $a$'s, then $q$ $b$'s, then $r$ $c$'s. Define a monomial $m(u;p,q,r)$ in $a,b,c$ as the product of all letters of $w$ such that the corresponding entries of $u$ are equal to 1. For example, if $u=101001, p=1, q=2, r=3, w=abbccc$, then $m(u;p,q,r)=abc$ because the first letter of $w$ is $a$, the third letter is $b$ and the sixth letter is $c$, and 1,3, 6 are the places in $u$ where $1$ occurs. Now define a polynomial $P(u;p,q,r)$ as the sum of all $m(u',w)$ for all cyclic shifts $u'$ of $u$. Let $S(u)$ be the array of all polynomials $P(u;p,q,r)$ for all $p,q,r$ with $p+q+r=n$ (indexed by the triples $(p,q,r)$).
Question Does the array $S(u)$ determine $u$ up to a cyclic shift?
The answer is "yes" for all $n\le 15$. A positive answer would imply a positive answer to this previous question for positive words. This question is a version of the present question when we only consider $q=1$. In that case, as Jyrki Lahtonen showed the answer is "no" even for $n=7$.