Let $F$ be a free group on $\{x_1,x_2,\cdots\}$, $w$ a word in $F$. Then $w$ is a finite expression in $x_i$'s and their inverses. By cancellation, $w$ can be reduced to simplest expression.
Let $w=x_{i_1}x_{i_2}\cdots x_{i_r}$ be a reduced word in $F$. It may happen that end letters can be inverses of each other, i.e. $x_{i_r}=x_{i_1}^{-1}$. In this case, $w$ can be written as $$w=x_{i_1} (x_{i_2}\cdots x_{i_{r-1}})x_{i_1}^{-1}.$$ Thus, a representative, namely $(x_{i_2}\cdots x_{i_{r-1}})$, of conjugacy class of $w$ is found which has length smaller than that of $w$. Continuing such process, we obtain an expression for $w$ as $$w=h(x_j x_k\cdots x_l)h^{-1},$$ in which
no cancellation can be done (or it is exactly same expression as beginning reduced form)
$x_j$ and $x_l$ are not inverses of each other.
This gives a minimum-length representative of conjugacy class of $w$ in $F$.
Question 1. Is this minimum-length conjugacy class representative uniquely determined? (I think, it is unique by the process described).
Question 2. If it is not uniquely determined, then is its length uniquely determined?
I confused with following sentence on Wikipedia (and so posted above questions):
Every word is conjugate to a cyclically reduced word. The cyclically reduced words are minimal-length representatives of the conjugacy classes in the free group. This representative is not uniquely determined, but it is unique up to cyclic shifts (since every cyclic shift is a conjugate element).