Two orbits are $G$- isomorphic iff their stabilizers are conjugate.

Suppose that a group $$G$$acts on a set $$A$$, we say $$A$$ is a $$G$$-set and two $$G$$- sets $$A$$ and $$B$$ are said to be$$G$$- isomorphic if there is a bijection $$f: A \mapsto B$$ such that $$g. f(a) = f(g.a)$$ for all $$g \in G, a\in A$$

My problem is: for any $$a,b \in A$$, orbits of $$a$$ and $$b$$ respectively, $$O_{a}, O_{b}$$ are isomorphic as G-sets if and only if there exists a $$g\in G$$ such that $$g \operatorname{stab}_{G}(a) g^{-1}= \operatorname{stab}_{G}(b)$$

I have trying to solve this for hours but unsuccessfully. My guess is to use that $$G$$-isomorphism condition to prove one inclusion of stabilizers and then use the same for $$f^{-1}$$ to prove the other inclusion of stabilizers.

Any help/hints that could help me solve this problem would be greatly appreciated, thanks!

As it is an "iff" question, it usually is easier to prove each direction separately. I'll try to describe the thought process.

First assume that $$g\operatorname{stab}_G(a)g^{-1}=\operatorname{stab}_G(b)$$ for some $$g$$. As we want to construct a bijection between $$O_a$$ and $$O_b$$ (and hope that it is a $$G$$-isomorphism), we might start by trying to "label" them by the same set. As $$O_a$$ and $$O_b$$ are orbits, we may use $$G$$ as a labeling set!

However, the labeling is not unique (if the action is not free): given $$x\in O_a$$, there may be two elements $$h_1,h_2\in G$$ such that $$x=h_1a=h_2b$$. In any case $$h_1h_2^{-1}\in\operatorname{stab}_G(a)$$, so $$gh_1h_2^{-1}g^{-1}\in\operatorname{stab}_G(b)$$ and so $$gh_1g^{-1}b=gh_2g^{-1}b$$.

Therefore the map $$\phi':O_a\to O_b$$, $$\phi'(ha)=ghg^{-1}b$$ is a well-defined bijection. However it is not a $$G$$-isomorphism, since instead it satisfies $$\phi'(hx)=ghg^{-1}\phi'(x).$$ We would be done if we had $$g=1$$. So recall that $$\operatorname{stab}_G(gb)=g\operatorname{stab}_G(b)g^{-1}=\operatorname{stab}_G(a)$$, and of course $$O_{gb}=O_b$$.

Applying the same process as above to $$gb$$ instead of $$b$$ yields a bijection $$\phi:O_a\to O_{gb}=O_b$$, $$\phi(ha)=hgb$$, satisfying $$\phi(hx)=h\phi(x)$$, and we're done.

The converse is similar: If $$\phi:O_a\to O_b$$ is a $$G$$-isomorphism, then $$\operatorname{stab}_G(a)=\operatorname{stab}_G(\phi(a))$$. Write $$\phi(a)=gb$$ for some $$g$$ (as it belongs to $$O_b$$) and use again that $$\operatorname{stab}_G(gb)=g\operatorname{stab}_G(b)g^{-1}$$.

Here is a hint. You can try to define a $$G$$-homomorphism $$\phi: O_a \to O_b$$ by sending $$a\mapsto b$$. Along with the condition $$\phi(g.a) = g.\phi(a)$$, this is enough to determine a unique map, if it exists. So, when is this map well-defined?