Say $(W,S)$ is a finite Coxeter group, such as a Weyl group (which satisfies an additional crystallographic condition).    Assume also that $W$ is irreducible.    Then it has a longest element $w_o$ relative to the given generating set $S$, which can be expressed as the $h/2$-power of a well-chosen Coxeter element when the Coxeter number $h$ is even (or a slightly modified expression if $h$ is odd).   

Assume now that $W$ is a Weyl group (the others being a little more complicated to study).   A standard fact is that $w_o = -1$ just when the type of $W$ in the classification is not ADE; more precisely, is different from $A_\ell (\ell \geq 2), D_\ell (\ell \geq 4), E_6$.   When $w_o =-1$, its centralizer is obviously $W$, but otherwise is a proper subgroup.

A recent question <a href="https://mathoverflow.net/questions/282727/type-c-n-weyl-group-contains-in-the-centralizer-of-the-longest-word-w-0-in?noredirect=1#comment697677_282727">*here*</a> suggests to me a possible uniform treatment of such centralizers in terms of *foldings* of Coxeter graphs.    But I'm not sure how far this is supported by the literature, or exactly how it might work for type $A_\ell$ with $\ell$ even --- then $W$ is the symmetric group $S_{\ell+1}$ and $h=\ell+1$ is odd. (I'm also unsure about the non-crystallographic types).  

Note that folding the graph of (say) type $E_6$ yields the graph $F_4$.   The Coxeter numbers are always the same when such foldings of ADE graphs occur, e.g., $h=12$ for both $E_6$ and $F_4$.  Moreover, $w_o$ is folded into the corresponding longest element $-1$ in accordance with the above expressions as powers of Coxeter elements if $h$ is even.   The question cited suggests that the "folded" Weyl group might always embed in $W$ as the precise centralizer of the original $w_o$.  

> Is this true, and is there a reference?   (Further, what can be said in the non-crystallographic case?)