Problem in permutation groups involving conjugates I have to find a permutation $a$ satisfying $ a xa^{-1}=y$  where $ x=(12) (34)$ and $y=(56) (13)$ 
My attempt in solving the problem was-
$$ a(12)(34)a^{-1}= a(12)(a^{-1}a)(34)a^{-1}= (a(12)a^{-1})(a(34)a^{-1})=(56)(13)$$
$$\implies (a(1) \ a(2)) (a(3) \ a(4))= (56)(13)$$
Comparing  both sides, $$ a(1)=5,\\ a(2)=6, \\  a(3)=1, \\ a(4)=3$$
Based on this information, I can write $a=\left(\begin{matrix}1 & 2 & 3 & 4 & 5 & 6\\ 5 & 6 & 1 & 3& 4& 2\end{matrix}\right)$,  which satisfies above four equations.
Is my solution correct  or not? If yes, thanks for confirming it and if not, then can someone please tell where I did go wrong?

Your solution is correct! One way to confirm this would be to write out
$$
a^{-1}=
\begin{pmatrix}
 5 & 6 & 1 & 3& 4& 2\\1 & 2 & 3 & 4 & 5 & 6
\end{pmatrix}
$$
and
$$
x=(12)(34)
$$
and
$$
a=
\begin{pmatrix}
 1 & 2 & 3 & 4 & 5 & 6\\5 & 6 & 1 & 3& 4& 2
\end{pmatrix}
$$
and simply follow the "route" of each element:
$$
5\overset{a^{-1}}{\longmapsto}1\overset{x}{\longmapsto}2\overset{a}{\longmapsto}6\\
6\overset{a^{-1}}{\longmapsto}2\overset{x}{\longmapsto}1\overset{a}{\longmapsto}5\\
1\overset{a^{-1}}{\longmapsto}3\overset{x}{\longmapsto}4\overset{a}{\longmapsto}3\\
3\overset{a^{-1}}{\longmapsto}4\overset{x}{\longmapsto}3\overset{a}{\longmapsto}1\\
4\overset{a^{-1}}{\longmapsto}5\overset{x}{\longmapsto}5\overset{a}{\longmapsto}4\\
2\overset{a^{-1}}{\longmapsto}6\overset{x}{\longmapsto}6\overset{a}{\longmapsto}2
$$
so $axa^{-1}$ acts exactly like $y$.
