Does every reflection generating set of a RA Coxeter group contain a conjugate of every standard generator? I am interested in understanding generating sets of right-angle Coxeter groups (RACGs) consisting of reflections. More precisely, let $(W,S)$ be a finite rank RACG, and write $R=\{wsw^{-1}\mid s\in S\;\textrm{and}\;w\in W\}$ for the set of reflections in $W$.
Question: Suppose $X\subset R$ generates $W$, then for each $s\in S$ is there $w\in W$ such that $wsw^{-1}\in X$? (Or indeed when is this the case for arbitrary Coxeter systems?)
This property doesn't hold for all Coxeter systems, for example for $W\cong\textrm{Dih}_5$, $\{s_1,s_2s_1s_2\}$ is a generating set, but in RACGs (or Even Coxeter groups more generally), distinct elements of $S$ are never conjugate so this kind of counter-example can't exist.
I tried searching the literature but couldn't find anything about this, however it could be relevant to note that RACGs are rigid, meaning that all Coxeter systems for a fixed RACG $W$ have isomorphic Coxeter-Dynkin diagrams. I did try to use the method of Pallavi Dani and Ivan Levcovitz, constructing folding sequences of cube complexes to prove the answer the question is yes, but couldn't quite make it work. Maybe there's some simple argument to answer the question?
 A: Apparently I'm not allowed to comment, so apologies for leaving this question as an answer: I don't understand what the problem is with the dihedral example: all reflections in the dihedral group of order $2 \times 5$ are conjugate, so in particular the generating set you gave contains conjugates of both the simple generates $s_1$ and $s_2$.
Edit: I put this in a comment but I should upgrade it to the answer: in a Coxeter group of rank $n$, every $n$-element generating set of reflections contains the same multiset of conjugacy classes as the set $S$ of simple reflections (whether or not it forms a simple system itself).  See Lemma 6.4 of https://arxiv.org/pdf/2110.14581.pdf by Patrick Wegener and Sophiane Yahiatene.
A: Let $(W,S)$ be a RACG of rank $n$, then it's abelianization is $W^{ab}=\mathbb{Z}_2^n=:\prod_{s\in S}\langle \overline{s}\mid\overline{s}^2\rangle$. If $X\subset W$ generates, then it's image in $W^{ab}$ must also generate. Suppose $X\subset R$, and consider $x=wsw^{-1}\in X$. It's image in $W^{ab}$ is $\overline{w}\,\overline{s}\,\overline{w}^{-1}=\overline{s}$, so if $X$ does not contain a conjugate of every element of $S$, then it's image in $W^{ab}$ wont generate, and hence $X$ cannot generate $W$. Thus the answer to question is yes.
More generally this argument also holds for even Coxeter systems, but in an arbitrary Coxeter system $(W,S)$ this argument allows one only to say the following: if $X\subset R$ generates $W$, then for each $s\in S$, there is $s'\in S$ which is connected to $s$ in the presentation diagram for $(W,S)$ by a path with all odd edge labels, such that $X$ contains an element conjugate to $s'$.
