A question about (unicity of certain cycles in a Cayley graph of a) symmetric group Let $S=\{(1,2),(1,2,3,\ldots,n),(1,2,3,\ldots,n)^{-1}=(1,n\ldots,2)\}$ be a subset of the symmetric group $S_n$. We know that $(1,2,\ldots,n)(1,2)=(2,3,\ldots,n)$, and thus $$[(1,2,\ldots,n)(1,2)]^{n-1}=(1,2,\ldots,n)\overbrace{(1,2)\cdots(1,2,\ldots,n)}^{2n-2}(1,2)=(1).$$ We want to know whether or not there exists another sequence of elements  $a_1,a_2,\ldots,a_{2n-4}\in S$ such that $$(12\ldots n)a_{2n-4}a_{2n-5}\cdots a_2a_1(12)=(1),$$ where  $a_{i+1}\neq a_i^{-1}$ for $i=0,1,2,\ldots,2n-4$ (putting $a_0=(12)$, $a_{2n-3}=(1,2,\ldots,n)$).
Equivalently, I want to ask if, in the cubic Cayley graph $Cay(S_n,S)$, there is a unique cycle of length $2(n-1)$ passing through $(1)$, $(1,2,\ldots,n)$ and $(1,2)$.
 A: The smallest $n$ for which there exist sequences as asked for is $n = 7$:


*

*$(1,2,3,4,5,6,7) \cdot (1,2) \cdot (1,7,6,5,4,3,2) \cdot (1,2)
     \cdot (1,2,3,4,5,6,7) \cdot (1,2) \cdot$ 
$(1,7,6,5,4,3,2) \cdot (1,2) \cdot (1,2,3,4,5,6,7) \cdot (1,2)
     \cdot (1,7,6,5,4,3,2) \cdot (1,2) = ()$, and

*$(1,2,3,4,5,6,7) \cdot (1,2,3,4,5,6,7) \cdot (1,2) \cdot (1,7,6,5,4,3,2)
     \cdot (1,7,6,5,4,3,2) \cdot$
$(1,2) \cdot (1,2,3,4,5,6,7) \cdot (1,2,3,4,5,6,7) \cdot (1,2) \cdot
     (1,7,6,5,4,3,2) \cdot$
$(1,7,6,5,4,3,2) \cdot (1,2) = ()$.
For $n = 8$ there is no such sequence other than the trivial one
mentioned in the question, for $n = 9$ there is one, for $n = 10$ there
are $18$, for $n = 11$ there are $5$ and for $n = 12$ there are $104$ such
sequences. This has been found with the following GAP function:
SearchXueyiSequences := function ( n )

  local  sequences, search, S;

  search := function ( sequence, a )

    local  b;

    if Length(sequence) = 2*n-3 then
      if sequence[2*n-3] <> (1,2) and Product(sequence)*(1,2) = () then
        Add(sequences,Concatenation(sequence,[(1,2)]));
      fi;
      return;
    fi;
    for b in Difference(S,[a^-1]) do
      search(Concatenation(sequence,[a]),b);
    od;    
  end;

  S := Set(GeneratorsAndInverses(SymmetricGroup(n)));
  sequences := [];
  search([],S[2]);
  sequences := Set(sequences);
  sequences := sequences{[2..Length(sequences)]}; # exclude trivial sequence
  return sequences; 
end;

A: For the case $n=3$, take $a_1 = (1,2)$ and $a_2 = (1,3,2)$ instead of 
$a_1 = (1,2,3)$ and $a_2 = (1,2)$.
A major motivation for posting this answer is to give Xueyi Huang an opportunity to exclude this case in one way or another if it is not really relevant.
Further edit on April 21:  As anticipated, Xueyi Huang has made a minor modification to the earlier form of the question and this answer is not a valid answer to the current form of the question.
