Let m1 and m2 belongs to the same orbit, that is m2=g∘m1 for some element g of a group G. Let us show that St(m2)=gSt(m1)g−1.
Let h∈St(m2). Then h∘m2=m2, and thus h∘(g∘m1)=g∘m1. It follows from definition of action that (hg)∘m1=g∘m1, and therefore, g−1∘((hg)∘m1)=g−1∘(g∘m1). Consequently, (g−1hg)∘m1=(g−1g)∘m1=e∘m1=m1. We conclude that g−1hg∈St(m1), and thus h∈gSt(m1)g−1. It follows that St(m2)⊂gSt(m1)g−1.
On the other hand, let h∈gSt(m1)g−1. Then h=gkg−1, where k∈St(m1), that is k∘m1=m1. It follows that h∘m2=(gkg−1)∘(g∘m1)=(gkg−1g)∘m1=(gk)∘m1=g∘(k∘m1)=g∘m1=m2 . We conclude that h∈St(m2), and therefore, gSt(m1)g−1⊂St(m2).
We concude that gSt(m1)g−1=St(m2), that is the stabilizers St(m1) and St(m2) are conjugate to each other.
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