Sylow Subgroups and Conjugation Suppose that $G$ is a finite group whose order is divisible by a prime $p$. Let $S$ be the set of Sylow $p$-subgroups of $G$; let $H$ be an element of $S$. $H$ acts on $S$ by conjugation. The fact that is the key to the proof of the Third Sylow Theorem is that there is only one $H$-orbit of order $1$. I was wondering if we could say more about this action.


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*I first thought that maybe there are only two $H$-orbits. But this turned out to be false in general. (I saw that it was false in the dihedral group $D_{5}$.)

*Next I checked whether all the orbits other than $\{H\}$ has order $H$. This is false in general, too. (This is false in $A_{5}$.)


After fiddling with some examples, I found that 

all the $H$-orbits except $\{H\}$ seem to have the same order.

But I can neither prove or disprove this claim. So here are my questions:


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*Is the above claim correct? If so, why?

*Is there any other things we can say about the $H$-orbits?
Any help is appreciated. Thanks in advance!

By computer calculation, the smallest counterexample is $S_3^2$ which has $9$ Sylow $2$-subgroups, and the action of one of them by conjugation has one fixed point, one orbit of length $4$, and two orbits of size $2$. 
