Conjugacy of simple system in a root system I'll set up the problem, then ask the question. 
Let $V$ be a finite dimension vector space over $\mathbb{R}$ and let $\Phi$ be a root system in $V$, i.e. (1) $\Phi \cap \mathbb{R} \alpha = \{-\alpha,\alpha\}$ for each $\alpha \in \Phi$ and (2) $s_\alpha \Phi = \Phi$ for each $\alpha \in \Phi$. A simple system $\Delta$ in a root system $\Phi$ is a subset of $\Phi$ such that $\Delta$ is a basis of $\text{span}_\mathbb{R} \Phi$ and each element of $\Phi$ is  a linear combination of $\Delta$, in which every coefficient has the same sign.
On page 10 of Humphreys' Reflection Groups and Coxeter Groups, he proves that any two positive (resp. simple) systems in $\Phi$ are conjugate under $W = $ the group genereated by the reflections $s_\alpha$, $\alpha \in \Phi$.
What does this theorem even mean? Really, what does "conjugate under $W$ mean"? If someone could explain this, then that would be nice.
 A: Notice that the Weyl group $W$ is created directly from your root system $\Phi$ (with no reference to a particular system of simple roots). Thus $W$ comes naturally from your root system.
A simple system on the other hand is proven to exist, but it is by no mean unique. Just like picking a basis for a vector space, one picks a simple system (among many choices). 
When doing linear algebra, there is always this question of whether something belongs to the vector space itself or if it is merely a mirage coming from your particular choice of basis (i.e. "Is [fill in the blank] coordinate independent?") For example, no matter how you choose coordinates, a linear operator has the same eigenvalues/determinant/trace. On the other hand, eigenvectors very much depend on the choice of coordinates (although they are related via changes of basis).
This same kind of issue needs to be addressed when working with root systems. Picking a simple system is like picking a basis. When we start studying the simple system, we would like to know if the things we are looking at depend on our particular simple system or not. 
The theorem you mention tells you that at the very least simple systems can be sent to each other via an element of the Weyl group. So if you notice property X when using a simple system, as long as property X is invariant under the Weyl group action, property X does not depend on the choice of simple system and thus is really a property of the root system itself.
Think of it this way: Vector Space/Change of Basis vs. Root System/Weyl Group
Change of basis = internal symmetry of a vector space
Weyl group = internal symmetry of a root system
A: You decide to study the root system $A_2 = \{ \vec{\mathbf{e}}_i - \vec{\mathbf{e}}_j : i\neq j \in \{1,2,3\} \}\subset \mathbb{R}^3$ this week.
You write out all the vectors $$\Phi=\{ (1,-1,0), (1,0,-1), (-1,1,0), (0,1,-1), (-1,0,1), (0,-1,1) \}$$ and check that $s_\alpha(\beta) \in \Phi$ for each of the $6 \times 6$ choices of $(\alpha,\beta)$. Actually that is a pain, and you realize you only need to check simple $\alpha$ and positive $\beta$. However, looking at $\Phi$ everything is symmetric!
You decide to make the right choice on Monday and let $\Delta = \{ (1,-1,0), (0,1,-1) \}$. You work out all sorts of important calculations, but before you are done you still have a few questions left.
On Tuesday you work on those questions, but you realize you forgot to bring your notes. You google for what $\Delta$ is supposed to be and find that $\Delta = \{ (0,-1,1), (-1,1,0) \}$. You manage to answer your remaining questions, and rejoice.
On Wednesday you remember all your notes and starting typing things up nicely, only to discover GASP! you used the wrong $\Delta$! Luckily, every possible choice of $\Delta$ is equivalent – you can just use a $s_\alpha$ to fix it.
In this case, to switch between Monday's and Tuesday's work, just apply $s_{(1,0,-1)}$ to all your calculations since $s_{(1,0,-1)}$ switches $(1,-1,0) \leftrightarrow (0,-1,1)$ and $(0,1,-1) \leftrightarrow (-1,1,0)$.
This means that Monday's $\Delta$ and Tuesday's $\Delta$ are conjugate under $s_{(1,0,-1)}$. To be conjugate under $W$, just means they are conjugate under some specific $w \in W$.
