Group theory, conjugation of permutations I have a past exam question that says...

Decompose the following permutations into a product of disjoint cycles. Are the two permutations conjugate?
$$\alpha= \begin{bmatrix}
   1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &9\\
   7 & 4 & 5 & 3 & 8 & 6 & 9 & 1 & 2\\
    \end{bmatrix}$$
and 
$$\beta= \begin{bmatrix}
   1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &9\\
   6 & 7 & 8 & 9 & 1 & 2 & 3 & 4 & 5\\
    \end{bmatrix}$$

I think that you have to find $x^{-1}\alpha x=\beta$, however I do not know how to find this $x$. Any help with this would be much appreciated. Thank you!
 A: Decomposition in cycle is fairly easy, and it yields
$$\alpha = (1\ 7\ 9\ 2\ 4\ 3\ 5\ 8)$$
$$\beta = (1\ 6\ 2\ 7\ 3\ 8\ 4\ 9\ 5)$$
Notice $6$ does not appear in the cycle decomposition of $\alpha$ (it would be a cycle of length 1, usually they are not written).
Two permutations of $S_n$ are conjugate iff they have the same cycle structure (see here): basically you only rename elements. Hence $\alpha$ and $\beta$ cannot be conjugate.

Where $\alpha$ and $\beta$ of same "cycle type", it would be easy to find a permutation to transform one into the other.
Say
$$\alpha=(1\ 7\ 4)(3\ 2\ 6\ 5)(8\ 9)$$
$$\beta=(3\ 4\ 7\ 6)(5\ 2\ 9)(1\ 8)$$
Then send each cycle to a cycle of same length: $1 \rightarrow 5$, $7 \rightarrow 2$...
So this one would do (it's not unique, just try to rotate any cycle)
$$x= \begin{bmatrix}
   1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &9\\
   5 & 4 & 3 & 9 & 6 & 7 & 2 & 1 & 8\\
\end{bmatrix}$$
Or $x=(1\ 5\ 6\ 7\ 2\ 4\ 9\ 8)$
Then $x\alpha x^{-1}=\beta$ if you use the convention $(xy)(i) = x(y(i))$, or $x^{-1}\alpha x=\beta$ with the other convention $(xy)(i)=y(x(i))$. In France it's usually the former, and in english-speaking countries, the latter.
For example, $x^{-1}({\color{red}{5}}) = 1, \alpha(1)=7, x(7)={\color{red}{2}}$, and $\beta({\color{red}{5}})={\color{red}{2}}$.

To answer mayi's comment below, let's try with $\alpha=(1\ 3)(4\ 7\ 6)$ and $\beta=(1\ 5)(2\ 6\ 4)$.
They have the same cycle structure, hence they are conjugate. The trick is you have to work in $S_7$, the group of permutations of $\{1,2,3,4,5,6,7\}$, so that $\alpha$ and $\beta$ share the same range. You may then use the permutation
$$\pi= \begin{bmatrix}
   1 & 2 & 3 & 4 & 5 & 6 & 7\\
   1 & - & 5 & 2 & - & 4 & 6\\
\end{bmatrix}$$
The images of $2$ and $5$ can be chosen freely, since they do not appear in $\alpha$, and the only remaining elements are $3$ and $7$. So, either $2\to3$ and $5\to7$, either $2\to7$ and $3\to5$. For instance:
$$\pi= \begin{bmatrix}
   1 & 2 & 3 & 4 & 5 & 6 & 7\\
   1 & 3 & 5 & 2 & 7 & 4 & 6\\
\end{bmatrix}$$
Or in cycle notation, $\pi=(2\ 3\ 5\ 7\ 6\ 4)$. Then, with the "french" convention of permutation composition, $\beta=\pi\alpha\pi^{-1}$, and with the "english" convention, $\beta=\pi^{-1}\alpha\pi$.
But $\pi$ is of course not the only possible permutation that transforms $\alpha$ into $\beta$. If you write $\beta=(1\ 5)(4\ 2\ 6)$, it's easy to see that $\pi=(2\ 7)(3\ 5)$ is also valid. To find your $x=(1\ 5\ 3)(2\ 7)$, write instead $\beta=(5\ 1)(4\ 2\ 6)$ (still with $\alpha=(1\ 3)(4\ 7\ 6)$), and
$$\pi= \begin{bmatrix}
   1 & 2 & 3 & 4 & 5 & 6 & 7\\
   5 & - & 1 & 4 & - & 6 & 2\\
\end{bmatrix}$$
Again, the images of $2$ and $5$ can be chosen freely. With $2\to7$ and $5\to3$, you get $\pi=(1\ 5\ 3)(2\ 7)$, your $x$. With the other choice, you would get $\pi=(1\ 5\ 7\ 2\ 3)$.
A: Easily we have
$$\alpha=(17924358)$$
and
$$\beta=(162738495)$$
Now assume that we can find $x$ such that
$$x^{-1}\alpha x=\beta$$
and since  $o(\beta)=9 $ and $o(\alpha)=8$ then
$$\beta^9={\rm{id}}=x^{-1}\alpha x\Rightarrow \alpha=\rm{id}$$
which is a contradiction.
A: Write down specifically the cycle decomposition of each permutation:
$$\begin{cases}\alpha=(17924358)\\{}\\\beta=(162738495)\end{cases}$$
Thus we have an $\;8-$ cycle and a $\;9-$ cycle...so are they conjugate?
