Slightly confused about the commutator subgroup

Let’s look at the group of all permutations of a set $left{ 1,2,3 right}$. There are exactly 6 elements in this group: $left( begin{matrix}
1 & 2 & 3 \
1 & 2 & 3 \
end{matrix} right)$, $left( begin{matrix}
1 & 2 & 3 \
1 & 3 & 2 \
end{matrix} right)$, $left( begin{matrix}
1 & 2 & 3 \
2 & 1 & 3 \
end{matrix} right)$, $left( begin{matrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{matrix} right)$, $left( begin{matrix}
1 & 2 & 3 \
3 & 1 & 2 \
end{matrix} right)$, $left( begin{matrix}
1 & 2 & 3 \
3 & 2 & 1 \
end{matrix} right)$. It is usually called the symmetric group $S_{3}$. Just out of curiosity I tried to find its commutator subgroup $left[ {{S}_{3}},{{S}_{3}} right]$. As a home assignment, I’ve already proven that the commutator subgroup is a normal subgroup, hence $forall xin {{S}_{3}}:xleft[ {{S}_{3}},{{S}_{3}} right]=left[ {{S}_{3}},{{S}_{3}} right]x$. Except for the neutral element $left( begin{matrix}
1 & 2 & 3 \
1 & 2 & 3 \
end{matrix} right)$, I couldn’t find a single permutation $p$ with the property: $forall xin {{S}_{3}}:xp=px$. For instance, $left( begin{matrix}
1 & 2 & 3 \
1 & 3 & 2 \
end{matrix} right)left( begin{matrix}
1 & 2 & 3 \
2 & 1 & 3 \
end{matrix} right)=left( begin{matrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{matrix} right)ne left( begin{matrix}
1 & 2 & 3 \
3 & 1 & 2 \
end{matrix} right)=left( begin{matrix}
1 & 2 & 3 \
2 & 1 & 3 \
end{matrix} right)left( begin{matrix}
1 & 2 & 3 \
1 & 3 & 2 \
end{matrix} right)$.

Am I missing something?