Request proof for proof for $x,y$ in the same orbit : if $y=gx,$ then $\mathrm{Stab}(y)=g\mathrm{Stab}(x)g^{-1}.$

Question

If $y,x$ lie in the same orbit, wrt action by the elements of the group $G,$ then need proof for : if $y=gx,$ then $\mathrm{Stab}(y)=g\mathrm{Stab}(x)g^{-1}.$

For example, have the below table (set members with colorings: $0000, 1111$ not shown, due to forming singleton orbits) for the action of the group of rotations of a tetrahedron, on the set $X$ of the $2$-colorings of the four vertices of the same.
Have the set $X$ size as $=2^4=16.$

Also, if take $x=1000, g= (234), y=(0100).$ Hence, $0010= (234)1000.$ And, $\mathrm{Stab}(1000)=\{e, (123), (132)\},$ $\mathrm{Stab}(0010)=\{e, (143), (134)\},$ $(234)(\{e, (123), (132)\})(432)$ $(234)(\{(432), (412),(13)(24)\}).$

Then, should get $\{e, (143), (134)\}=(234)\{e, (123), (132)\}(432).$ Applying composition of permutations, on the rhs, from right to left, get: $(234)\{e, (123), (132)\}(432)$ $=(234)\{(432),(124), (13)(24)\} =\{e, (134), (143)\}$

\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \text{S. No.} & e & (234) & (432) & (123) & (132) & (134) & (143)& (124) & (142) & (12)(34) & (13)(24)& (14)(23) & \mathcal {O}\\ \hline 1& 0001 & 0001 = 1 & 0001=1 & 0010=2 & 0100 =4& 0100 =4& 1000 = 8& 0010= 2 & 1000 = 4& 0010=2 & 0100=4 & 1000=8 & \{1,2,4,8\}=\mathcal O_1 \\ 2 & 0010 & 0100=4 & 1000 =8& 0100=4 & 0001=1 & 0010 =2& 0010 =2& 1000=8 & 0001=1 & 0001 =1& 1000 =8& 0100=4 & \{1,2,4,8\}=\mathcal O_1 \\ 3 & 0011 & 0101=5 & 1001=9 & 0110=6 & 0101=5 & 0110 =6& 1010=10 & 1010=10 & 1001=9 & 0011 =3& 1100=12 & 1010= 10 &\{3,5,6,9,10,12\}=\mathcal O_2 \\ 4 & 0100 & 1000=8 & 0010=2 & 0001=1 & 0010=2 & 1000=8 & 0001=1 & 0100=4 & 0100=4 & 1000=8 & 0001=1 & 0010= 2 &\{1,2,4,8\}=\mathcal O_1 \\ 5 & 0101 & 1001=9 & 0011=3 & 0011=3 & 0110=6 & 1100 & 1001=9 & 0110=6 & 1100=12 & 1010=10 & 0101=5 & 1010=10 & \{3,5,6, 9, 10, 12\}=\mathcal O_2 \\ 6 & 0110 & 1100 =12& 1010=10 & 0101=5 & 0011=3 & 1010=10 & 0011=3 & 1100=12 & 0101=5 & 1001=9 &1001=9 & 0110=6 & \{3,5,6,9,10,12\} =\mathcal O_2 \\ 7 & 0111 & 1101=13 & 1011=11 & 0111=7 & 0111=7 & 1110=14 & 1011=11 & 1110=14 & 1101=13 & 1011=11 &1101=13 & 1110=14&\{7, 11,13,14\}=\mathcal O_3 \\ 8 & 1000 & 0010=2 & 0100=4 & 1000=8 & 1000=8 & 0001=1 & 0100=4 & 0001=1 & 0010=2 & 0100 =4& 0010=2 & 0001=2&\{1,2,4,8\}=\mathcal O_1 \\ 9 & 1001 & 0011=3 & 0101=5 & 1010=6 & 1100=12 & 0101=5 & 1100=12 & 0011=3 & 1010=10 & 0110=6 & 0110 = 6& 1001=9&\{3,5,6,9,10,12\}=\mathcal O_2 \\ 10 & 1010 & 0110=6 & 1100=12 & 1100=12 & 1001=9 & 0011=3 & 0110 =6& 1001=9 & 0011=3 & 0101=5 &1010=10 & 0101=5&\{3,5,6,9,10,12\}=\mathcal O_2 \\ 11 & 1011 & 0111=7 & 1101=13 & 1110=14 & 1101=13 & 0111=7 & 1110=14 & 1011=11 & 1011=11 & 0111=7 & 1110=14 & 1101=13&\{7,11,13,14\}=\mathcal O_3 \\ 12 & 1100 & 1010 =10& 0110=6 & 1001=9& 1010=10 & 1001=9 & 0101=5 & 0101=5 & 0110=6 & 1100=12 & 0011=3 & 0011=3&\{3,5,6,9,10,12\}=\mathcal O_2 \\ 13 & 1101 & 1011=11 & 0111=7 & 1011=11& 1110 =14&1101=13 & 1101 =13& 0111=7 & 1110 =14& 1110=14 & 0111 =7& 1011=11&\{7,11,13,14\}=\mathcal O_3 \\ 14 & 1110 & 1110 =14& 1110=14 & 1101=13 & 1011=11 & 1011 =11& 0111=7 & 1101 =13& 0111=7 & 1101=13 & 1011=11 & 0111 =7&\{7,11,13,14\}=\mathcal O_3 \\ \end{array}

Answer 1

Let $h\in stab(gx) $,

$$\begin{align} hgx=gx &\implies g^{-1}hgx=x \\ &\implies g^{-1}hg \in stab(x) \\ &\implies h\in g Stab(x) g^{-1} \end{align}$$

So $$Stab(y) \subset gStab(x) g^{-1}$$ In the same way, you can prove the reverse inclusion by noting that it is equivalent to $$Stab(x) \subset g^{-1}Stab(y)g $$

Answer 2

Directly from the set-builder notation (namely without the double inclusion): \begin{alignat}{1} \operatorname{Stab}(y) &= \{g'\in G\mid g'y=y\} \\ &= \{g'\in G\mid g'gx=gx\} \\ &= \{g'\in G\mid g^{-1}g'gx=x\} \\ \tag1 \end{alignat} Now, call $g'':=g^{-1}g'g$; then, $g':=gg''g^{-1}$, which plugged into $(1)$ yields: \begin{alignat}{1} \operatorname{Stab}(y) &= \{gg''g^{-1}\in G\mid g''x=x\} \\ &= g\{g''\in G\mid g''x=x\}g^{-1} \\ &= g\operatorname{Stab}(x)g^{-1} \\ \tag2 \end{alignat}