Consider
$$\bf C_k = \left[\begin{array}{cccc}0&\bf I_{k-1}\\1&0\end{array}\right]$$
if $$\bf P = {\bf TC_kT}^{-1}$$
and $\bf T$ also a permutation matrix. In other words: will this hold:
$$\nexists i,l<k : ({\bf P}^l)_{ii} = 1$$
I would guess so, but I have no proof. I'm about to make an application about it so it would be good to know if it holds or not.
On
If I understand your question, what you are doing is nothing but conjugating some permutation in ${\cal S}_k$ by another permutation.
But conjugation in the symmetric group preserves the type (i.e. the set of lengths of the disjoint cycles making up the permutation) and since ${\bf C}_k$ is a cycle of length $k$ so must be ${\bf T}{\bf C}_k{\bf T}^{-1}$.
But if $\pi$ is a cycle of length $k$ in ${\cal S}_k$, $\pi(j)\neq j$ for all $j=1,...,k$ which is what your question amounts to.
I think I have finally understood this question. The trace of a conjugate of a matrix equals the trace of the matrix. The matrix $C_k$ has trace zero, so $P=TC_k T^{-1}$ has trace zero. As $P$ is a permutation matrix, with only entries $0$ and $1$ all the elements on the diagonal must be zero, since they add to zero.