Depth of the in-tree of $s$ under $q \mapsto q s q^{-1}$ on $n$-cycles

Fix an $n$-cycle $s$ and map every $n$-cycle $q$ to its conjugate $D(q) = q s q^{-1}$, which is the same as reading the one-line word $(q(0), \ldots, q(n-1))$ back as a cycle. Iterating $D$ turns the $(n-1)!$ $n$-cycles into a functional graph. Its only fixed point is $s$, and the cycles that eventually reach $s$ form a tree feeding into it. How deep is that tree? Exactly $\varphi(n)$ cycles map directly onto $s$, and the tree stays shallow - depth 1 - unless $8 \mid n$ or $p^2 \mid n$ for an odd prime $p$, which is the Hull-Dobell threshold for the existence of a full-period non-translation affine map on $\mathbb{Z}/n$. Past it the depth is $p^{e-1}$ for $n = p^e$ with $p$ odd, $2^{e-1} - 1$ for $n = 2^e$, and for general $n$ the largest of these over the prime powers dividing $n$.

Opus 4.8 constructed a branch of the stated depth, giving a lower bound, and believed it had a matching upper bound; that proof was wrong and the statement stayed a conjecture. FABLE 5 later proved it. In the author's summary of the method: "The proof turns conjugation, near $s$, into base-$p$ arithmetic." A cycle near the fixed point splits into a coarse base permutation and a vector of carries in $\mathbb{Z}/p$, $D$ acts on the carries by the carrying of ordinary base-$p$ addition, and the depth comes out as the nilpotency length of a shift difference - exactly that for odd $p$, one less for $p = 2$. The single missing carry that odd primes absorb and $2$ cannot is what produces the two-branch answer. A companion survey paper covers the rest of the graph: the other periodic orbits, congruences on basin sizes, and a cyclic-sieving count. The depth theorem is the substantive part.