So this is from Charles C. Pinter's "A Book of Abstract Algebra"- specifically, it's from the second chapter on permutations. The question is:
Let $\alpha_1$ and $\alpha_2$ be cycles of the same length. Let $\beta_1$ and $\beta_2$ be cycles of the same length. Let $\alpha_1$ and $\beta_1$ be disjoint, and let $\alpha_2$ and $\beta_2$ be disjoint. [Prove that] There is a permutation, $\pi\in S_n$, such that $\alpha_1\beta_1=\pi\alpha_2\beta_2\pi^{-1}$.
That's the question.
But this is actually the last of a five part question, sooooo here are the previous parts - the parts build on each other and it looks like part 4 (which can be found near the bottom) is of greatest relevance but I can't quite see how to use it here.
Part 1 was:
Let $\alpha=(a_1,...,a_s)$ be a cycle and let $\pi$ be a permutation in $S_n$. Then $\pi\alpha\pi^{-1}$ is the cycle $(\pi(\alpha_1),...,\pi(\alpha_s))$.
(a solution can be found here Proof for conjugate cycles)
Part 2 was:
Conclude from part 1: Any two cycles of the same length are conjugates of each other.
The idea here is not too hard to formulate and some formalization is given in a hint which suggests defining a permutation, $\pi$, as sending any element not in either cycle to itself, sending all of the elements of one of the cycles, denoted by A, to the other, denoted by B. Finally, define $\pi$ over B-A as some bijection between B-A and A-B. Now, it is easy to show that $\pi$ is a permutation and by part 1, that's enough.
Parts 3 and 4 are straightforward.
Part 3:
If $\alpha$ and $\beta$ are disjoint cycles, then $\pi\alpha\pi^{-1}$ and $\pi\beta\pi^{-1}$ are disjoint cycles.
(this follows directly from part 1)
And Part 4:
Let $\sigma$ be a product $\alpha_1...\alpha_t$ of t disjoint cycles of lengths $l_1,...,l_t$, respectively. Then $\pi\sigma\pi^{-1}$ is also a product of t disjoint cycles of lengths $l_1,...,l_t$.
(an easy generalization of part 3)
In the fifth part, I see how one can find a $\pi$ such that $\alpha_1=\pi\alpha_2\pi^{-1}$. I also see how one can find a $\pi$ such that $\beta_1=\pi\beta_2\pi^{-1}$ (we can get this via an immediate application of part 2). What I can't see is how we can ensure that these two $\pi$'s will be the same (which is what the question seems to be asking). I also can't see how to use Part 4 here as, on its own, it doesn't seem enough- it looks to me like we would need, in addition, to show that the act of taking a conjugate could be made 'onto' in a sense (and I don't get how to do that).
Any help appreciated.
Let
$\alpha_1=(\alpha^1_1,\alpha^1_2,\dots,\alpha^1_n),\alpha_2=(\alpha^2_1,\alpha^2_2,\dots,\alpha^2_n),$
$\beta_1=(\beta^1_1,\beta^1_2,\dots,\beta^1_m),\beta_2=(\beta^2_1,\beta^2_2,\dots,\beta^2_m)$.
Let $\pi$ be any permutation which satisfies $\pi(\alpha^2_i)=\alpha^1_i$ and $\pi(\beta^2_j)=\beta^1_j$ for all $1\le i \le n$ and $1\le j \le m$. Then $\pi\alpha_2\beta_2\pi^{-1}=\pi\alpha_2\pi^{-1}\pi\beta_2\pi^{-1}=\alpha_1\beta_1$, using part (1) of your problem.
Edit: I see now that the crux of your question is why such a permutation $\pi$ exists. For any given pair of lists of $k$ distinct numbers $x_1,\dots,x_k$ and $y_1,\dots,y_k$, there exists a permutation which satisfies $\pi(x_i)=y_i$. In fact, there are $(n-k)!$ choices for $\pi$. To choose $\pi$, let $z_1,\dots,z_{n-k}$ be a list of the numbers not present in $x_1,\dots,x_k$, then for each $j=1,2,\dots,n-k$ in order, set $\pi(z_j)$ to be any number unequal to $y_1,\dots,y_k,z_1,z_2,\dots,z_{j-1}$.