What is the length of the anti-diagonal permutation in the centralizer of that element? Consider the symmetric group $S_n$. Write the transposition $\sigma_i := (i, i+1)$.  In case $n = 2m$, write $$t = \sigma_m, ~~~s_i := \sigma_{m-i}\sigma_{m+i}~~ (1\leq i\leq m-1)$$ In case $n = 2m+1$, write $$t= \sigma_m\sigma_{m+1}\sigma_m,~~~s_i = \sigma_{m-i}\sigma_{m+i+1}~~~(1\leq i\leq m-1)$$
Now consider the element, $\alpha \in S_n$, represented by an $n \times n$ anti-diagonal matrix.  The elements $s_1,...,s_{m-1}, t$ generate the subgroup fixed by conjugation $$\sigma \mapsto \alpha^{-1} \sigma \alpha.$$  If $t_0 = t$, $t_i = s_i t_{i-1}s_i$, it is not difficult to see that $\alpha$ can be written as the product $\alpha = t_0t_1...t_{m-1}$.  Thus the "length" of $\alpha$ with respect to the elements $t, s_1,...,s_{m-1}$ is not greater than  $1 +3+5+...= m^2$.  Why is it precisely $m^2$? That is, why can't $\alpha$ be written as a shorter word in the elements $t,s_1,...,s_{m-1}$?
 A: I only write the answer for the case of even $n$, but the odd case should work in a similar way.
What you have there is a Coxeter group of type B and its longest element. If you google that, I'm sure you will find information on why this longest element has exactly length $m^2$. For example, looking at this Wikipedia page, we get

If the Coxeter group is finite then the length of w0 is the number of the positive roots.

The number of positive roots is the same as the number of reflections, i.e. the number of conjugates of your generators $t$ and $s_i$ in your group. The set of conjugates of these generators is exactly
$$\{(i,j)(m-i,m-j) \mid 1 \leq i < j \leq m\} \cup \{(i,m-i) \mid 1 \leq i \leq m\}.$$
I'm not sure how much theory of Coxeter groups you need to answer your question, it might be possible to do it brutally by hand without using any of it. But if you are interested in these kinds of groups, I would suggest to look them up, the length of the longest element is one of the rather basic results you will find.
