Are $s_1, s_3$ in $S_4$ conjugate? Let $S_4$ be the symmetric group consisting of permutations over $\{1,2,3,4\}$. Let  $s_1 = \left( \begin{matrix} 1 & 2 & 3 & 4 \\ 2 & 1 & 3 & 4 \end{matrix} \right)$ and $s_3 = \left( \begin{matrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 4 & 3 \end{matrix} \right)$. Is there some $g \in S_4$ such that $g s_1 g^{-1} = s_3$? Thank you very much.
 A: Notice that in cycle notation $s_1=(1 \ 2)$ and $s_2=(3 \ 4)$. Thus, for any $g \in S_4$ you will have $g s_1 g^{-1} = (g(1) \ g(2))$. So what happens if you choose $g$ such that $g(1)=3$ and $g(2)=4$?
A: Here is the general approach to that:
Let $a = (a_1,a_2,\ldots, a_k)$ and $b = (b_1,b_2,\ldots b_m)$ be two cycles in $S_n$ for some $n \geq k,m$. Then $a,b$ are conjugated if and only if $k = m$. In this case, any element of $S_n$ that maps $a_i$ to $b_i$ for all $i$ will give the desired result. More general, let $\phi \in S_n$. Then we define the cycle counter (or cycle type) of $\phi$ to be $(c_1,c_2,\ldots, c_n)$, where 
$c_i =$ number of cycles of length $i$ in the disjoint cycle representation of $\phi$.
Now extending the above result, two permutations are conjugated if and only if they have the same cycle type.
In your case, you have $s_1 = (1,2)$ and $s_2 = (3,4)$ and thus for example the permutation $(1,3)(2,4)$ will do the trick.
Remark: Depending on your definition of conjugation, if you put the inverse left or right, you might need to use the inverse, that is not $a_i \mapsto b_i$ but rather $b_i \mapsto a_i$.
