Let $G=S_n$, $X=S_n$. $G$ operates on $X$ by conjugation: $g.x=gxg^{-1}$. Question: What do the orbits look like and apply your result on $S_5$.
I already could calculate for $x=(n_{1,1}\ldots n_{1,m_1})\ldots(n_{l,1}\ldots n_{l,m_l})$ in cycle notation that $$gxg^{-1}=(g(n_{1,1})\ldots g(n_{1,m_1}))\ldots(g(n_{l,1})\ldots g(n_{l,m_l})).$$ So every orbit looks like$$Gx=\{(g(n_{1,1})\ldots g(n_{1,m_1}))\ldots(g(n_{l,1})\ldots g(n_{l,m_l})): g\in G\}.$$
Still I don't know how I can apply this on $S_5$? Can someone give me a hint? :)
EDIT: I gave it a thought again. I was assuming that $x$ is already a disjoint cycle. Now by the bijectivity of $g$ $$(g(n_{1,1})\ldots g(n_{1,m_1}))\ldots(g(n_{l,1})\ldots g(n_{l,m_l}))$$ is a disjoint cycle, as well.
The other way round: For any disjoint cycle $$y=(z_{1,1}\ldots z_{1,m_1})\ldots(z_{l,1}\ldots z_{l,m_l})$$ of same lenght of $x$ , there exists an $g\in S_n$, namely $g(n_{ij}):=z_{ij}$, such that $gxg^{-1}=y$. $g\in S_n$, since the cycles of $x$ and $y$ are disjoint.
Conclusion: $$Gx= \{(g(n_{1,1})\ldots g(n_{1,m_1}))\ldots(g(n_{l,1})\ldots g(n_{l,m_l})): g\in G\}=\{(n_{1,1}\ldots n_{1,m_1})\ldots(n_{l,1}\ldots n_{l,m_l}): n_{ij}\in\{1,\ldots,n\}\text{ disjoint }\}$$
Since $X$ is disjoint union of it's orbits, we have:
$$S_5=G(12)\cup G(123)\cup G(1234)\cup G(12345)\cup G(12)(345)$$
which is of course $S_5$ since every element can be written in a disjoint cycle of such length of $(12),(123),(1234), (12345)$ or $(12)(345)$.
Hope this is correct now. :)
Any element of $S_n$ can be decomposed into disjoint cycles. What you have shown is that the "cycle type" does not change under conjugation by $S_n$. For example, $(12)(345) \in S_5$ and its conjugacy class consists of all those elements that are a product of a $2$-cycle and a $3$-cycle that are disjoint.
By the way, you missed the computation for $G((12)(34))$ but otherwise your work is correct.