Let $G$ be a group acting on a set $A$. Assume that stabilizer of any element $a, b\in A$ are isomorphic to each other, $Stab(a)\cong Stab(b)$. Would that imply that there is only one orbit?
My approach was if they are isomorphic, then every element of $A$ is fixed under the action of $G$. It means that stabilizer only consists of identity element. Therefore there is only one orbit. But I think this approach is far away from being complete.
Let $V=\Bbb Z_2\times\Bbb Z_2$ be the Klein-4 group. Define the subgroups $X=\Bbb Z_2\times\{0\}$ and $Y=\{0\}\times\Bbb Z_2$. Let $U$ and $V$ be copies of $X$ and $Y$ which $V$ acts on componentwise. The stabilizer of both points in $U$ is $Y$, and the stabilizer of both points in $V$ is $X$, and we find that $X\cong Y$ are isomorphic, but $U$ and $V$ are distinct orbits.
Note that stabilizers being isomorphic is not really connected to the group action - it is usually an irrelevant coincidence. The relationship between stabilizers which is actually relevant to the group action is being conjugate subgroups, because $\mathrm{Stab}(gx)=g\mathrm{Stab}(x)g^{-1}$.
The stronger question, then, is if two elements have conjugate stabilizers then do they lie in the same orbit? No: you can just let $G$ act on two or more distinct, disjoint copies of the same orbit, then you get conjugate stabilizers but as many orbits as you want. (Note my previous example is not an instance of this, since $V$ is abelian no distinct subgroups can be conjugate.)
The true statement is that two elements having conjugate stabilizers imply their orbits are isomorphic as $G$-sets (that is, there exists a equivariant aka intertwining bijection between them).