Necessary and sufficient for $\operatorname{orb}(x)=\operatorname{orb}(y) \iff \operatorname{Stab}(x)=g\operatorname{Stab}(y)g^{-1}$

Question

Are orbits equal if and only if stabilizers are conjugate?

You may get some insights from the link above.

My Question: What is the necessary and sufficient condition for the above statement to be true?

I can prove with no difficulty that same orbits implies conjugate stabilizer .For the converse, if $G$ is non trivial groups, is it strong enough to show the converse is true? I guess it probably not working with only the condition but i couldn't get a counter example.

Any hints, counter example, ideas or related information would be appreciated.

Answer 1

If $O(x)=O(y)$ then $Stab(x)$ and $Stab(y)$ are conjugate.

Since they are in same orbit, there exist a $g\in G$ such that $g*x=y$.

Now, let $r\in Stab(y)$ then $$r*y=y$$ $$r*(gx)=gx$$ $$(g^{-1}rg)*x=x$$ so $(g^{-1}rg)\in Stab(x)\implies g^{-1}Stab(y)g \leq Stab(x)$ to show other inclusion you can fallow similiar way then you are done.

Converse is not true. If $Stab(x)$ ans $Stab(y)$ are conjugate, you can not conclude that $O(x)=O(y)$. But you can say that $|O(x)|=|O(y)|$ by ortbir-stabilizer theorem.

Let $G=Z_2=\{0,1\}$ and $\omega=\{1,2,a_1,a_2 \}$ and the zero of $G$ fixes every element and $1_G$

$$1_G:1\to 2$$ $$1_G:2\to 1$$ $$1_G:a_1\to a_2$$ $$1_G:a_2\to a_1$$

You can check that $Stab(1)=Stab(a_1)$ so they are also conjugate to each other but $O(1)=\{1,2\}$ and $O(a_1)=\{a_1,a_2\}$.