A subtle error in a "change of variable" in a sum Let $G$ be a finite group of order $n$, and let $E$ be a finite set. Let $\star$ be an action of $G$ on $E$.
Suppose that $G \star x_1,..., G \star x_m$ are the distinct orbits of elements in $E$. 
For each $i \in \{1,..,m\}$, define:
$$P_i = \bigcup_{\large t \in G \star x_i} G_{t} \times \{ t \}$$
Where $G_t$ denotes the stabilizer of $t$.
It is required to prove that for each $i$, $|P_i| = n = |G|$.
(Some background: this is part of an exercise whose aim is to establish Burnside's lemma)
I realize that my method is incorrect and I already know how to establish that $|P_i| = n$, but I am trying to understand what is wrong with the following (initial) reasoning of mine:
$$|P_i| = \left| \bigcup_{\large t \in G \star x_i } G_t \times \{t\} \right| = \sum_{\large t \in G \star x_i} \left| G_t \times \{t\}  \right| = \sum_{\large t \in G \star x_i} \left| G_t\right| = \sum_{a \in G} \left| G_{a \star x_i} \right| \\ = \sum_{a \in G} \frac{|G|}{|G \star (a \star x_i)|} = \sum_{a \in G}\frac{|G|}{|G \star x_i|} = \frac{|G|}{|G \star x_i|} \sum_{a \in G} 1 = \frac{|G|^2}{|G \star x_i|} = |G| \times |G_{x_i}|$$
Can someone point out my mistake?
My thoughts are that I went wrong when I "changed the variable" (hence the title), but I still can't figure out why would that be invalid.

The mistake is in the equality
$$\sum_{\large t \in G \star x_i} \left| G_t\right| = \sum_{a \in G} \left| G_{a \star x_i} \right|$$at the end of the first line; the problem is that you want to write every $t\in G\star x_i$ as $a\star x_i$ for some $a\in G$. You can do this, except there is generally more than one such $a$, so you are counting too many times the same $G_t$. In fact,$$a\star x_i = a'\star x_i \Leftrightarrow (a')^{-1 }a\in G_{x_i},$$so every $|G_t|$ is counted exactly $|G_{x_i}|$ times, which is why your final answer has the extra factor $|G_{x_i}|$.
