This is question 8.26 from Dan Saracino's abstract algebra. I have showed that, for every $h \in S_n$, $h \circ (x_1,...,x_r) \circ h^{-1} = (h(x_1),...,h(x_r))$.
But part b) is as follows,
I tried doing the only if direction. My attempt : Suppose $f_1 = f_{n_1} \circ \cdot \cdot \cdot \circ f_{n_r}$ and $f_2 = g_{n_1} \circ \cdot \cdot \cdot \circ g_{n_r}$, where f, g are disjoint.
Then I defined $h = \begin{pmatrix} f_{n_1} & ... & f_{n_r}\\ g_{n_1} & ... & g_{n_r} \end{pmatrix}$, so $h \circ f_1 \circ h^{-1} = g_{n_1} \circ \cdot \cdot \cdot \circ g_{n_r} = f_2$
But for the other direction I am stuck.
Please correct me if my attempt is wrong, I need some help thank you.
The if part is much easier.
$f_2=h \circ f_1 \circ h^{-1}$
$ = h \circ f_{n_1} \circ \cdot \cdot \cdot \circ f_{n_r} \circ h^{-1}$
$ = h \circ f_{n_1} \circ h^{-1} \circ h\circ f_{n_2} \circ h^{-1} \circ h \circ f_{n_3} \cdot \cdot \cdot \circ f_{n_r} \circ h^{-1}$
$=g_{n_1} \circ \cdot \cdot \cdot \circ g_{n_r}$,
where $g_{n_i}=h \circ f_{n_i} \circ h^{-1}$. Hence $g_{n_i}$ and $f_{n_i}$ have same length. So $f_1$ and $f_2$ have same number of r-cycles for every r.