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The permutation character of $H$ on $X$ (where $\lvert X\rvert\ge2$) has the form $1_G+\chi_X$, where $\chi_X$ is irreducible of degree $\lvert X\rvert-1$. Thus if $\lvert X\rvert\ne\lvert Y\rvert$, then the scalar product $(1_G+\chi_X, 1_G+\chi_Y)$ is $1$. By the theorem which is attributed to (but which is not due to) Burnside we see that $H$ is transitive on $X\times Y$. This implies that $H_x$ is transitive on $Y$.

Examples of this situation are the Mathieu group $M_{11}$ or $\text{PSL}_2(11)$, which both have doubly transitive actions on $11$ and $12$ points.

For these two examples, we actually can argue directly. If $\lvert Y\rvert=12$, then $11$ divides the order of $H_y$, so $H_y$ acts transitively on the set $X$ of size $11$.

The permutation character of $H$ on $X$ (where $\lvert X\rvert\ge2$) has the form $1_G+\chi_X$, where $\chi_X$ is irreducible of degree $\lvert X\rvert-1$. Thus if $\lvert X\rvert\ne\lvert Y\rvert$, then the scalar product $(1_G+\chi_X, 1_G+\chi_Y)$ is $1$. By the theorem which is attributed to (but which is not due to) Burnside we see that $H$ is transitive on $X\times Y$. This implies that $H_x$ is transitive on $Y$.

A slight variant of the argument is as follows: Let $\pi_X$ be the permutation character of $H$ on $X$, and $(\cdot,\cdot)$ be the scalar product of characters. Note that \begin{equation} 0\le(\pi_X-\pi_Y, \pi_X-\pi_Y)=(\pi_X, \pi_X)+(\pi_Y, \pi_Y)-2(\pi_X, \pi_Y). \end{equation} Double transitivity of $H$ on $X$ and $Y$ means $(\pi_X, \pi_X)=(\pi_Y, \pi_Y)=2$, so $(\pi_X, \pi_Y)\le2$ with equality if and only if $\pi_X=\pi_Y$. Thus if $\lvert X\rvert\ne\lvert Y\rvert$, then $(\pi_X, \pi_Y)=1$, and we finish as above.

Examples of this situation are the Mathieu group $M_{11}$ or $\text{PSL}_2(11)$, which both have doubly transitive actions on $11$ and $12$ points.

For these two examples, we actually can argue directly. If $\lvert Y\rvert=12$, then $11$ divides the order of $H_y$, so $H_y$ acts transitively on the set $X$ of size $11$.

The point stabilizer $H$ of the Mathieu group $M_{11}$ in its $2$-transitive action on $12$ points acts transitively in the doubly transitive action of $M_{11}$ on $11$ points. This follows because $11$ divides $\lvert H\rvert$. (Or from the fact that the scalar product of the two permutation characters of degrees $11$ and $12$ is $1$.)

Another example is $\text{PSL}_2(11)$ in its actions on $11$ and $12$ points.

The permutation character of $H$ on $X$ (where $\lvert X\rvert\ge2$) has the form $1_G+\chi_X$, where $\chi_X$ is irreducible of degree $\lvert X\rvert-1$. Thus if $\lvert X\rvert\ne\lvert Y\rvert$, then the scalar product $(1_G+\chi_X, 1_G+\chi_Y)$ is $1$. By the theorem which is attributed to (but which is not due to) Burnside we see that $H$ is transitive on $X\times Y$. This implies that $H_x$ is transitive on $Y$.

Examples of this situation are the Mathieu group $M_{11}$ or $\text{PSL}_2(11)$, which both have doubly transitive actions on $11$ and $12$ points.

For these two examples, we actually can argue directly. If $\lvert Y\rvert=12$, then $11$ divides the order of $H_y$, so $H_y$ acts transitively on the set $X$ of size $11$.

The point stabilizer $H$ of the Mathieu group $M_{11}$ in its $2$-transitive action on $12$ points acts transitively in the doubly transitive action of $M_{11}$ on $11$ points. This follows because $11$ divides $\lvert H\rvert$. (Or from the fact that the scalar product of the two permutation characters of degrees $11$ and $12$ is $1$.)

The point stabilizer $H$ of the Mathieu group $M_{11}$ in its $2$-transitive action on $12$ points acts transitively in the doubly transitive action of $M_{11}$ on $11$ points. This follows because $11$ divides $\lvert H\rvert$. (Or from the fact that the scalar product of the two permutation characters of degrees $11$ and $12$ is $1$.)

Another example is $\text{PSL}_2(11)$ in its actions on $11$ and $12$ points.