On the product of two permutations (and its conjugates)

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.