How can I fully understand what are conjugacy classes are in groups?
I know the definition, that $a$ and $b$ are conjugate if $gag^{-1}=b$ for some $g\in G$. But what is the intuition? Using a multiplication table (all the possible multiplications), how can I understand what a conjugacy class is?
For example, on Wikipedia you can see this table:
How can I relate the idea of a conjugacy class with this table?
There is one way to understand conjugacy from the perspective of linear algebra, using change-of-basis.
Conjugacy can also be interpreted as a "measure of departure" from being a commutative group. Given two elements $g,h$ of a group $G$ let us look at the element $h^{-1}gh$.
We modify $g$ by multiplying by $h$ on the right. To get back the $g$ we should multiply this result by $h^{-1}$ again on the right. But instead we multiplied it by $h^{-1}$ on the left. That is we used the same ingredients "but mixed them in different order". If it gives back $g$ well and good then these elements are said to commute. If it does not give back $g$, we interpret it as something "close to $g$, a conjugate of $g$. In an abelian group if two elements are close enough to be conjuagate they are in fact same. In a non-abelain group we consider $h^{-1}gh$ for various elements $h$ as giving use many elements that are close to $g$, or 'similar to" $g$ (Linear algebra usus this terminology.)
If an element has many elements similar/conjugate to it (other than itself) it is a measure of failure of commutativity. The larger the number of elements conjugate to it, more non-commutative that element is.
Conjugate elements have same order, and have same behaviour in many respects. (In matrices conjugate elements have same trace, determinant and eigenvalues).
It's not easy to see conjugacy in a group table, but try this:
Look at the $f_v$ row, and the $f_h$ column. $f_d$ is in both of them, and it's in the $r_1$ column of the $f_v$ row, and the $r_1$ row of the $f_h$ column. That tells you $f_vr_1=r_1f_h$ (since both equal $f_d$), which says $f_v=r_1f_hr_1^{-1}$, that is, $f_v$ and $f_h$ are conjugate.
In general, $a$ and $b$ are conjugate if there are elements $x$ and $y$ such that $x$ is in the $a$ row and the $y$ column, and also in the $b$ column and the $y$ row. If there are no such $x$ and $y$, then $a$ and $b$ aren't conjugate.
Of course, this is nothing but a restatement of the definition of conjugacy --- I'm just trying to restate it in terms of the group table, as requested.
Let me use an analogy from linear algebra. In linear algebra two matrices $X, Y$ are conjugate iff. there exists a matrix $M$ such that $X = MYM^{-1}$. In linear algebra, this has a specific interpretation: $X$ is the change of basis of $Y$ where $M$ is the matrix representing the transformation of the basis of $Y$ to the basis of $X$.
It turns out that this interpretation transfers exactly into group theory. Let us consider the symmetry group $S_n$ and let us express the elements of $S_n$ in cycle notation. let $x = (abc)(de)$, $y = (cba)(de)$, and $g = (ca)$. The element $x$ can be considered as a "change of basis" (with the change being switching $c$ for $a$ $y$) from $y$ with the element representing the "basis transformation" (substitution in this case) being $g$.
