How to understand orbit of size $1$ in this case I am a self-studying beginner in group theory, so please bear with this question which could have some simple answers. Given a $p$-group $G$ for some prime $p$, let $H$ be a subgroup of $G$. Let $X$ be the set of all conjugates of $H$.
Now, $H$ acts on $X$ by conjugation. I read that there are at least $p$ orbits of size $1$ in $X$.
One example of an orbit with size $1$ is $\{H\} \in X$. This example follows since $aHa^{-1}=H$ for any $a \in H$ since $H$ is a subgroup, and we have $\text{Orb}(H)=H$.
But I read that since $p$ is prime, that there are at least $p-1$ other orbits of size $1$. So there should be another orbit $gHg^{-1} \neq H$ of size $1$ in $X$.
What I don't understand is how $gHg^{-1}$ could be of size $1$ under the action of $H$. Shouldn't this mean that $\text{Orb}(gHg^{-1})=\{agHg^{-1}a^{-1} | a \in H\}$ and $\text{Orb}(gHg^{-1})$ may not necessarily be equal to $gHg^{-1}$. However, it should have size $1$, which means that $\text{Orb}(gHg^{-1})$ should in fact be equal to $gHg^{-1}$.
For reference, this result came from Rotman's Theorem 4.6, where no extra conditions were imposed on $H$ and $G$ except that $H$ is a subgroup of the $p$-group $G$ ... What am I missing here?

The first thing to note is that if $|X| = 1$ then we will not have $p-1$ other orbits so we will also need to assume $|X| \gt 1$.
We will use these two properties of orbits to prove our statement:

*

*Orbits are disjoint and their union is the entire set $X$ (this should be easy to see).


*The orbit size divides group order (this is proven in the Orbit-stabilizer theorem)
By property (1) we have that $$|X| = \sum_{Y \in \mathcal{O}} |Y|$$ where $\mathcal{O}$ is the set containing all the orbits of the action. Now we split $\mathcal{O}$ into two disjoint subsets: $\mathcal{O'}$ and $\mathcal{O''}$ where $\mathcal{O'}$ is the set of all orbits of size $1$ and $\mathcal{O''}$ is the set of all orbits of size greater than $1$. This means $$|X| = \sum_{Y' \in \mathcal{O'}} |Y'| + \sum_{Y'' \in \mathcal{O'}} |Y''| = |\mathcal{O'}| + \sum_{Y'' \in \mathcal{O'}} |Y''|$$ since $|Y'| = 1$. By property (2) we know that $|Y''|$ divides $|X| = p^n$ and $|Y''| > 1$ which means that $|Y''| = p^k$ where $k > 1$ which means $p$ divides $|Y''|$. We can view $X$ as an orbit where the group action is conjugation by the group $G$. This means that $|X|$ divides $|G| = p^n$. Since $|X| > 1$ we have that $p$ divides $|X|$. Since $|X| =  |\mathcal{O'}| + \sum_{Y'' \in \mathcal{O'}} |Y''|$, $p$ also has to divide $|\mathcal{O'}|$ which means $|\mathcal{O'}| = pm$ for some $m \gt 1$ which means $|\mathcal{O'}| \geq p$ which is what we were trying to prove.
