Two cyclically reduced words in a free group are conjugate if and only if one is a cyclic permutation of the other

If $u=pq$ literally and $v=qp$, then $p^{-1}up=p^{-1}pqp$ freely reduces to $qp=v$, so every cyclic permutation of $u$ is conjugate to $u$.

For the converse, suppose $v=t^{-1}ut$ in the reduced-word group and take $t$ reduced. If $t=\varepsilon$, then $u=v$ as elements of the underlying set of reduced words in [L1], which is a cyclic permutation obtained by taking an empty prefix.

Assume the converse holds for conjugators shorter than a nonempty reduced word $t=a t'$, where $a$ is its first letter.

If neither seam in the literal word $t^{-1}ut$ cancels, that word is reduced and begins with the inverse of its last letter, so [F1] says it is not cyclically reduced; but by [L1] this reduced word is the group product $t^{-1}ut$ and hence equals the cyclically reduced word $v$, a contradiction. Thus at least one of the two seams cancels.

If the right seam cancels, write $u=u'a^{-1}$; then $t^{-1}ut$ freely reduces to $(t')^{-1}(a^{-1}u')t'$, where $a^{-1}u'$ is a cyclic permutation of $u$. If the left seam cancels, write $u=au'$; then it freely reduces to $(t')^{-1}(u'a)t'$, where $u'a$ is a cyclic permutation of $u$. In either case the shifted word is cyclically reduced and the conjugator $t'$ is shorter.

Apply [L3] to the property that the converse holds for every conjugator of length at most $n$. The induction hypothesis makes $v$ a cyclic permutation of the shifted word in step 3.1; cyclic permutations compose, so $v$ is a cyclic permutation of $u$.

If one of $u,v$ is empty, conjugacy forces both to be the identity element, which is the empty word in the reduced-word group of [L1]. Combining this boundary with steps 1.1 and 4.1 proves both directions in the reduced-word model, and [L2] transports the criterion to every free group on $X$.