Let \(X\) be a \(G\)-set with group action \(G \times X \to X\). (1) Prove that if \(x_1, x_2 \in X\) belong to the same \(G\)-orbit (\(x_2 = g \cdot x_1\) for some \(g \in G\)), then their stabilizer subgroups \(\operatorname{Stab}_G(x_1)\) and \(\operatorname{Stab}_G(x_2)\) are conjugate in \(G\): \[\operatorname{Stab}_G(g \cdot x_1) = g \operatorname{Stab}_G(x_1) g^{-1}.\] (2) Note that if \(x_1, x_2\) lie in different orbits, their stabilizers need not be conjugate.
Solution
For \(h\in G\), \[ \begin{aligned} h\in\operatorname{Stab}_G(gx) &\iff h(gx)=gx\\ &\iff (g^{-1}hg)x=x\\ &\iff g^{-1}hg\in\operatorname{Stab}_G(x)\\ &\iff h\in g\operatorname{Stab}_G(x)g^{-1}. \end{aligned} \] Hence \[ \operatorname{Stab}_G(gx)=g\operatorname{Stab}_G(x)g^{-1}. \] Thus points in the same orbit have conjugate stabilizers.
For different orbits this can fail. For example, let \(S_3\) act on \[ \{1,2,3\}\sqcup\{*\} \] in the usual way on the first orbit and trivially on \(*\). Then \[ \operatorname{Stab}(1)\cong C_2, \qquad \operatorname{Stab}(*)=S_3, \] so the stabilizers are not conjugate.