How does showing $\tau(i) = j \implies a\tau a^{-1}(a(i)) = a(j)$ demonstrate that $a\tau a^{-1}$ has the same cycle type as $\tau$? I am completely new to group theory. My class is beginning to learn about centralizers, which, if I am not mistaken, is the set of all elements $x$ in a group $G$ that commute with a specified element $a$ in $G$. To help us calculate the cardinality of the centralizer of an element in $S_5$, my teacher has introduced the orbit-stabilizer theorem $|O_x| \cdot |G_x| = |G|$ (I still do not understand what an orbit or stabilizer refers to). He states that the cardinality of the centralizer of an element in $S_5$ is $|G_x|$ (because I am still confused as to what a stabilizer is, I am not sure how this is true) and that $|O_x|$ is the "conjugacy class" (I do not remember the exact terminology my teacher employed), so we can calculate $|G_x|$ by finding $\frac{|G|}{|O_x|}$.
To lead us to finding $|O_x|$, my teacher has stated that we have to show an equivalence between the "conjugacy class" and the cycle types of $S_5$; to do so, we must show  $\tau(i) = j \implies a\tau a^{-1}(a(i)) = a(j)$ for cycles $a, \tau$.
I understand that $a\tau a^{-1}(a(i)) = a\tau (a^{-1}a)(i) = a\tau(i) = a(j)$. However, how does showing this implication demonstrates that $a\tau a^{-1}$ has the same cycle type as $\tau$. Additionally, how does this relate to the supposed equivalence between the "conjugacy class" and cycle types of $S_5$?
Thanks in advance!
 A: Here we have the group $G$ acting on itself by conjugation. That is, for $g \in G$ and $h \in G$ have that $\alpha : G \times G \to G$ by $\alpha(g,h) = ghg^{-1}$.
Now for a given $h \in G$. The stabilizer $G_h$ are all $g$ such that $ghg^{-1} = h$ multiplying on the right by $g$ we get that $gh = hg$. Thus, the stabilizers are exactly the centralizers.
Now let $i_1 \to i_2 \to \ldots \to i_k \to i_1$ be a cycle in the cycle decomposition of $\tau$. Then by the above $a(i_1) \to a(i_2) \to \ldots \to a(i_k) \to a(i_1)$ is a cycle in the cycle decomposition of $a\tau a^{-1}$. Thus, they have the same same cycle type.
Now to complete the proof, we have to show that if $\tau_1$ and $\tau_2$ are two permutations with the same cycle type, then there exists a permutation $a$ such that $a\tau_1 a^{-1} = \tau_2$. But this also follows from your computation (just match cycles in the cycle types and define $a$ using that).
Thus, this tells us that the orbit of a permutation is exactly all permutations with the same cycle type. Thus, we can compute the size of the by orbit by counting. Thus, we get the size of the stabilizers and hence the size of centralizer.
