the orbit of a root under operations of irreducible crystallographic group? Suppose we have an irreducible crystallographic coxeter group G acting in a vector space V, how can we show that the orbit of an element in its root system $\Delta$ is the set of the roots of the same length?
I have shown this for the ADE types of groups, which is quite easy since if G is an ADE group, then for any element $r_i, r_j\in\Delta, \exists T\in G: Tr_i=r_j$. Thus the orbit of any single element in the root system of a group of the ADE type is just the entire root system itself, which are of course roots of the same length because ADE groups have roots of the same length.
But how would I show this for the non ADE types? ie. $F_4, G_2, B_n$
 A: Ok, so there is a simple and uniform (in terms of crystallographic root systems) way to answer your question without doing case by case work (at least not too much). I am not going to repeat it here, since it is very nicely presented in the following blog entry Properties of Irreducible Root Systems III. However, I feel I can contribute a bit more, since you reminded me of a question I was wondering about in the past, and because of your post, I finally went to the trouble of answering it.
So, what you are asking can be rephrased as "How many conjugacy classes of reflections are there in a Weyl group?". Of course, this makes sense in any (finite) reflection group and that is what I had been asking myself in the past. The answer turns out to be at most two, again!
Actually, there is a method to figure out whether any two involutions of the reflection group are conjugate, just by looking at the associated Dynkin diagram (see for instance Chapter VIII -Conjugacy Classes- of the very beautiful "Reflection Groups and Invariant Theory" by Richard Kane).
For reflections, this becomes particularly nice: First, recall that any reflection is conjugate to one of the simple reflections. Now, two simple reflections that are adjacent in the Coxeter diagram are conjugate if and only if the edge that connects them has an odd label.
The forward implication is trivial. If $(s_is_j)^{2k+1}=1$, then $(s_is_j)^ks_i(s_js_i)^k=s_j$, that is $s_i$ and $s_j$ are conjugate. The other implication is less straight forward and comes down to showing that any two conjugate involutions can be shown to be conjugate via a sequence of elementary equivalences. Put together, these mean that two simple reflections are conjugate exactly when you can walk from one to the other (in the Dynkin diagram), without ever stepping on an edge with an even label.
Now, a simple check on the classification of the Coxeter-Dynkin diagrams, will tell you that there can only be one even-labelled edge, so there are at most two conjugacy classes of reflections! Notice, that for this, as well as your original question, we only need the trivial implication from above (but also the classification...).
