Let $G < S_n$ be a permutation group of degree $n$, and let $G^{(i)},0 < i\leq n$, be the pointwise stabilizer of $\{1,2.., i\}$ in $G$. We set $G^{(0)}= G$. For $0 < i\leq n$, let $U_i$ be a complete right transveral for $G^{(i)}$ in $G^{(i-1)}$. If $\psi$ and $\psi'$ are distinct coset representatives in $U_i$, then
$i^{\psi} \neq i^{\psi'}$
But $\psi, \psi'$ both fix $\{1,..i-1\}$. If $i^{\psi} = i^{\psi'}$, still $\psi$ and $\psi'$ can be in distinct coset representatives in $U_i$.
Would anyone eplain why being $\psi$ and $\psi'$ in distinct coset representatives in $U_i$ infers $i^{\psi} \neq i^{\psi'}$?
Suppose $i^{\psi} = i^{\psi'}$, with $\psi, \psi' \in G^{(i-1)}$.
Then $i^{\psi' \psi^{-1}} = i$, so $\psi' \psi^{-1} \in G^{(i-1)}$ fixes also $i$, and thus $\psi' \psi^{-1} \in G^{(i)}$.
It follows that $G^{(i)} \psi = G^{(i)} \psi'$.