A question about (unicity of certain cycles in a Cayley graph of a) symmetric group

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;

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.