Every two permutations of order $2$ in $S_4$ are conjugate I was trying to solve the following question:

Prove or disprove: every two permutations of order $2$ in $S_4$ are conjugate.

I tried to disprove it: $\sigma_{1}=(2,3)$ and $\sigma_{2}=(1,2)(3,4)$ so $\sigma_1,\sigma_2\in S_4$. Also
$$  \sigma_{1}\cdot\sigma_{1}=id \Rightarrow o(\sigma_{1})=2\\
 \sigma_{2}\cdot\sigma_{2}=id \Rightarrow o(\sigma_{2})=2. $$
Let's check the conjugate:
$$ \tau^{-1}(1,2)(3,4)\tau=(2,3)\Leftrightarrow(\tau(1),\tau(2))(\tau(3),(\tau(4))=(2,3).$$
They have different structure, so they are not conjugate.
Is my proof valid?
Also what is the difference between $ \tau^{-1}(1,2)(3,4)\tau$ and $ \tau (1,2)(3,4)\tau^{-1}$? Are they both equal to $(\tau(1),\tau(2))(\tau(3),(\tau(4))$?
Edit: How should my answer change if the order of the two permutations is $3$? I think that in that case the the theorem is right. Consider $\sigma_1=(a,b,c)$ and $\sigma_2=(x,y,z)$ then $\tau^{-1}(a,b,c)\tau=(x,y,z)$ so we get $(\tau(a),\tau(b),\tau(c))=(x,y,z)$. But what now?
 A: It is fine. In general, there is a theorem that two permutations in $S_n$ are conjugate if and only if they have the same cycle structure. You can take it as a good exercise to try to prove it. So $(23)$ and $(12)(34)$ have no chance to be conjugate. On the other hand $(12)$ and $(23)$ are conjugate, we know it without even checking. This is a very important theorem. 
Edit: note that $\tau^{-1}(a,b,c)\tau=(\tau^{-1}(a),\tau^{-1}(b),\tau^{-1}(c))$, not $(\tau(a),\tau(b),\tau(c))$. Composition of permutations is done from right to left, just like any composition of functions. 
A: Two cycles $\sigma,\tau$ in $S_n$ in general are conjugate iff they have the same cycle structure. In this case, because $\sigma,\tau$ have order $2$, disjoint cycles commute and $n$-cycles always have order $n$, we must have that each cycle in the cycle decomposition of $\sigma,\tau$ has order dividing $2$ (though this is not always a sufficient criterion). The reason the statement is false in this case is because we can exhibit two different cycle shapes satisfying this criterion: $(\cdot,\cdot)(\cdot)(\cdot)$ and $(\cdot,\cdot)(\cdot,\cdot)$ which as you observe, both have order $2$.
But suppose we were not working in $S_4$ but $S_3$. Then because $\sigma,\tau$ are order $2$, which is prime, we must have a cycle of length $2$. Then we must have a $1$-cycle left over, so the only cycle shape is $(\cdot,\cdot)(\cdot)$. So indeed, any two cycles of order $2$ in $S_3$ are conjugate!
A: Obviously false by the fact that any two permutations that are conjugate have the same cycle type.   
For example, $(12)$ and $(12)(34)$ both have order two.
The converse of the above fact is also true:  any two permutations of the same cycle type are conjugate in $S_n$.  Thus if you consider elements of order $3$, since they must be $3$-cycles (reason?), they are conjugate. 
