Question about permutation cycles

[xentity1x]

If A is a cycle, and A=(1 4 5) (2 3 6). Is there a B such that BAB^-1=A^2. I found
A^2=(1 5 4) (2 6 3), but I'm not really sure where to go from there.

 

[Dick]

You know they are conjugate because they have the same cycle structure. If you need to find a B, then looking at your results, I'd say you want a B that reverses 5 and 4 and reverses 6 and 3. Can you think of one?