The period of the decimal expression for $1/n$ is at most $n-1$, not $n$. Equality holds if and only if $n$ is prime and $10$ is a primitive root modulo $n$.

Notice that for a solution, $y$ cannot be prime or prime power. Indeed, your equation is equivalent to demanding that $(x+1)(x^2-x+1)/y^3$ be a positive integer less than $x$. If $p$ is the only prime divisor of $y$, then since $\gcd(x+1, x^2-x+1)$ is $1$ or $3$, we would get $p=3$ (otherwise $(x+1)(x^2-x+1)/y^3$ would be at least $x+1$). However, $x^2 - x + 1$ is never divisible by $9$, so this cannot happen.

@CarloBeenakker I assume it's the inverse transpose. (So there is an implicit assumption that I-X is invertible, I guess.) But I don't see how this can be true for arbitrary upper triangular matrices: it's even false for diagonal matrices, as far as I can see. Perhaps the OP meant strictly upper/lower triangular.

@YCor Oh, it's only now that I see that you described $F$ and not $G$ as $\mathbf{Q}^{(c)} \times \mathbf{Q}/\mathbf{Z}$. So the question remains what $G$ is (as a topological group) if we equip $F$ with the discrete topology.

(By the way, what I wrote about the group being independent of the choice of the field is not correct; I had the subgroup in mind of elements starting with $a_0=1$, which is independent of the field.)

@YCor We do want to take the topology into account arising from the projective limit (but with $F$ equipped with the discrete topology), so then $\mathbf{Q}_p$ is not just the same as $\mathbf{R}$. In addition, I don't really see where the factor $\mathbf{Q}/\mathbf{Z}$ comes from. (Notice that we consider a fixed prime $p$).

@YCor Thanks — but what is $\mathbf{Q}^{(c)}$? Could you expand (and perhaps post this as an answer)? We do not take the topology of $\mathbf{C}^*$ into account, but I don't see how it plays a role: the resulting group is independent of the choice of the field provided it has $p^n$-th roots for all $n \geq 0$ and characteristic not equal to $p$. Could you explain what you mean? Thanks!