Question about permutation cycles We have: $\alpha = (a_1a_2 \cdots a_r), \beta=(b_1b_2\cdots b_r)\in S_n$ ($\alpha,\beta$ are strange cycles) 
How can we find $f\in S_n$ s.t.:
$$\beta=f\alpha f^{-1}\;\;\;?$$
Thank you!
(The answer should be as multiply of strange permutation cycles...) 
strange = disjoint
 A: The key to answering this question is knowing that 
$$f\alpha f^{-1} = (f(a_1)\, \dots\, f(a_r)).$$
To see this, note that any $b \in \{1, \dots, n\}$ is equal to $f(a)$ for some $a$. If $a \notin \{a_1, \dots, a_r\}$, then 
$$(f\alpha f^{-1})(f(a)) = f(\alpha(a)) = f(a).$$ If $a = a_i$, then $$(f\alpha f^{-1})(f(a_i)) = f(\alpha(a_i)) = f(a_{i+1})$$ where $a_{r+1}$ should be interpreted as $a_1$. Therefore, 
$$f\alpha f^{-1} = (f(a_1)\, \dots\, f(a_r)).$$
With this fact at your disposal, you want to find $f$ such that
$$f\alpha f^{-1} = (f(a_1)\, \dots\, f(a_r)) = (b_1\, \dots\, b_r).$$
As Derek Holt suggested in the comments, you can take $f(a_i) = b_i$ and extend to a bijection. Note, this is not the only such $f$.
A: You could approach this from a linear algebra perspective. You first would write the permutation matrices corresponding to $\alpha$ and $\beta$, lets call them $A$ and $B$ respectively. You are now searching for the matrix $F$ such that
$B = F A F^{-1}$.
To find $F$ you can follow the steps in this answer. $F$ would therefore map to a linear function $f$.
