Find a permutation

For $x = (12)(34)$ and $y = (56)(13),$ find a permutation $a$ such that $a^{-1}xa = y$.

I written $a^{-1}xa = y$ as $xa = ay$ thus $(12)(34)a = a(56)(13)$ but I can't find the $a$?

Conjugation affects cycles like so: $\sigma(a_1~a_2~\cdots~a_r)\sigma^{-1}=(\sigma(a_1)~\sigma(a_2)~\cdots~\sigma(a_r))$.
• In particular: (1) do you understand what I've said so far? (2) can you extrapolate to figure out how conjugation explicitly affects a product of (disjoint) cycles? (3) so then (rearranging $a^{-1}xa=y$ $\Leftrightarrow x=aya^{-1}$) what does $a(56)(13)a^{-1}$ look like? and finally (4) what can you select $a$ to be to achieve $(12)(34)$? – anon Nov 13 '13 at 4:32
• $a^{-1}(12)(34)a = (56)(13)$. I am stuck in finding such an a. – user104235 Nov 13 '13 at 4:32