Convoluted permutation conjugate proof but is it sound?

Question

Spent a long time trying to prove that if two permutations have the same cycle structure, then they are conjugate.

Wondering if this proof is sound, I understand there are other proofs (simpler ones) but I don't really understand them. I've managed to come up with this proof though I'm not fully convinced would appreciate if someone could let me know.

I've managed to prove that any two $k$ cycles are conjugate, from now on I will just refer to this as lemma 1. We also know that every element in $S_n$ is a product of disjoint cycles and disjoint cycles commute in $S_n$.

Now consider two permutations $ \alpha, \tau \in S_n$ such that $\alpha = a_1 a_2\cdots a_k$ and $\tau = b_1 b_2\cdots b_k$ where $a_i,b_i$ are cycles of length $i$. Now from lemma 1 we know that any two k cycles are conjugate thus one can write $\rho_1a_1\rho_1^{-1} = b_1$ and $\rho_2a_2\rho_2^{-1} = b_2$ all the way up to $\rho_ka_k\rho_k^{-1} = b_k $.

If one considers the following product keeping in mind that all these elements commute as they belong to $S_n$: \begin{align}(\rho_1a_1\rho_1^{-1})(\rho_2a_2\rho_2^{-1}) \cdots (\rho_ka_k\rho_k^{-1}) &= (\rho_1\rho_2\cdots\rho_k)(a_1a_2\cdots a_3)(\rho_1\rho_2\cdots\rho_k)^{-1} \\ &= \alpha\sigma\alpha^{-1} \\ &= b_1b_2\cdots b_k \\ &= \tau \end{align}

Where $\alpha$ is the product of the $\rho$ elements.

Thus the permutations $\sigma$ and $\tau$ are conjugate

Answer 1

In your question you've mentioned that you are familiar with the fact that

$$\pi(i_1\cdots i_k)\pi^{-1} = (\pi(i_1)\cdots\pi(i_k))$$

You can derive the general case with a trick (also keep in mind that $\mathrm{id} = \pi\pi^{-1}=\pi^{-1}\pi$):

\begin{align} \delta & = \pi\sigma\pi^{-1}=\pi(i_1\cdots i_k)(j_1\cdots j_n) \cdots (f_1\cdots f_m)\pi^{-1} \\ & =\pi(i_1\cdots i_k)\pi^{-1}\pi(j_1\cdots j_n) \pi^{-1}\pi\cdots\pi(f_1\cdots f_m) \pi^{-1} \\ & = (\pi(i_1) \cdots \pi(i_k))(\pi(j_1)\cdots\pi(j_n))\cdots(\pi(f_1) \cdots \pi(f_m)) \end{align}

Hope that I've answered the question for you!

UPD:
Sorry, but the proof you've added is not correct: you are assuming that $ρ_i$ and $a_i$ are disjoint so they commute, but that's not necessarily the case. Consider an example: $\pi\sigma\pi^{-1}=(12)(123)(12)$ (as $(12)=(12)^{-1}$). You can check yourself that the fact still holds.