Let $G$ be a group acting transitively on the space $X$. Let $x\in X$ and write $X=G\cdot x=G/H$.
Let $A$ be a subgroup of $G$ and consider the orbit $Y:=A\cdot x$. Suppose that $g\cdot Y\cap Y\not=\varnothing$ for some $g\in G$. Is it necessary that $g\cdot Y=Y$?
My attempt: Suppose that $ga_1\cdot x=a_2\cdot x$ for some $a_1,a_2\in A$ then $a_2^{-1}ga_1\in H$ so $g=a_2ha_1^{-1}$. Now for any $a\in A$ we have $ga\cdot x = a_2ha_1^{-1}\cdot x=a_2h\cdot x_1$ for some $x_1\in Y$. the question becomes now is $h\cdot x_1\in Y$?
The answer is no.
Here's an intuitive counterexample. Suppose $G$ is the group of rigid motions of the plane $X=\mathbb{R}^2$, in which case the stabilizer of (say) the origin is the rotation group $H=\mathrm{SO}(2)$. Let $A\cong\mathbb{R}$ be the set of translations along some line, so $Y$ is any parallel line. Then if $g\in G$ is a rotation (not $0^{\circ}$ or $180^{\circ}$), the line $gY$ intersects the line $Y$, but they are not parallel hence not the same line.
The smallest counterexample will occur with $G=S_3$ acting on $X=\{1,2,3\}$. Since it is so small I will let you try your hand at finding $A$ and $Y$ and $g$.