What does the proposition of stabilizers really tell us? I am trying to understand the following

Proposition: Let $G$ be acting on a non empty set $T$,
Suppose $t_{1}=g \star t \in \operatorname{orb}(t)$, then $\operatorname{stab}(t_{1})=g \operatorname{stab}(t) g^{-1}$

So what is this really saying, that for some $t$ , we have the $\operatorname{orb}(t)= \{ g \star t : g \in G \}$, and we take one of these and call it $t_{1}$ and it is associated with a specific $g$?
then what stabilizes $t_{1}$ ie $\operatorname{stab}(t_{1})=\{g \in G : g \star t_{1} = t_{1})$ is the same as the conjugation of all of $\operatorname{stab}(t)$ with the prior $g$?
I dont know if I really understand it, or what it is saying. Can anyone help to clarify this? Thanks
 A: The proposition could be restated as the following:
Let $G$ act on a non-empty set $T$. If $t_1,t\in T$ are in the same orbit, then the stabilizers of $t_1$ and $t$ are conjugate.
Can you see why $t_1\in orb(t) \iff orb(t_1)=orb(t)$?
A: What you said is pretty much correct. Start with some element $t \in T$. The orbit of $t$ is the set of all elements of $T$ which can be "reached" from $t$ via the group action, in other words, all elements of the form $g \star t$.
Now take an arbitrary element $t_1$ from the orbit of $t$. It must be of the form $t_1 = g \star t$ for some $g \in G$. What elements of $G$ should stabilize $t_1$? Well, if we start with $g \star t$, and apply $g^{-1}$, we get $g^{-1} \star (g \star t) = (g^{-1}g)\star t = t$. Now we know what stabilizes $t$: it is precisely $stab(t)$. So if we start with $g \star t$ and apply $g^{-1}$ and then apply any element $s \in stab(t)$, the result is $s \star t = t$. If we now apply $g$, we get back $g \star t$.
So think of $g\ stab(t)\ g^{-1}$ as the composition of three actions (applied from right to left): first an action by $g^{-1}$ which moves $g \star t$ to $t$, then an action by $s \in stab(t)$, which fixes $t$, then an action by $g$, which moves $t$ back to $g \star t$, which is where we started. In other words, $g\ stab(t)\ g^{-1}$ stabilizes $g \star t$.
