Let $G$ be a finite group and $A$ be a subgroup of $G$. Suppose $A^x$ and $A^y$ are two conjugates of $A$ such that $A$, $A^x$ and $A^y$ are not equal. Do we always have $|A\cap A^x|=|A\cap A^y|$?
I tried to find a counterexample in $S_4$ but failed. Does this hold for any finite group and its subgroups? Any help is appreciated! Thanks!
On
A counterexample can be found for example in $G=S_3 \times S_3$: put $A=\{((1),(1)),((1),(12)),((23),(1)),((23),(12))\}$. Then $A$ is a Sylow $2$-subgroup of $G$. And so are $B=\{((1),(1)),((1),(13)),((23),(1)),((23),(13))\}$ and $C=\{((1),(1)),((13),(13)),((13),(1)),((1),(13))\}$. Of course, by Sylow Theory all $2$-Sylow subgroups are conjugated. But here $A \cap B=\{((1),(1)),((23),(1))\}$, while $A\cap C=\{((1),(1))\}$. So, $|A \cap B|=2$ and $|A \cap C|=1$.
View $G$ as a transitive group acting on the set of cosets $\Omega=[G:G_\alpha]$, where we write $A=G_\alpha$. Then \begin{equation*} G_\alpha\cap G_\alpha^x=G_{(\alpha,\alpha^x)} \end{equation*} is the point-wise stabiliser of $\{\alpha,\alpha^x\}$. Consider the action of $G_\alpha$ on $\Omega\setminus\{G_\alpha\}$. The stabiliser of $\alpha^x$ in $G_\alpha$ is exactly $G_{(\alpha,\alpha^x)}$, the order of which is \begin{equation*} G_{(\alpha,\alpha^x)}=\frac{|G_\alpha|}{|(\alpha^x)^{G_\alpha}|}. \end{equation*}
If the length of orbits $|(\alpha^x)^{G_\alpha}|$ and $|(\alpha^y)^{G_\alpha}|$ containing $\alpha^x$ and $\alpha^y$ respectively are different then $|G_{(\alpha,\alpha^x)}|\ne|G_{(\alpha,\alpha^y)}|$. Usually there are several subdegrees in a transitive action, and so suitable $x$ and $y$ are easily obtained. That means, your equation can only hold in very special cases.