I need to find and prove a necessary and sufficient condition when two cycles $x=(x_{1}x_{2}\cdots x_{m})$ and $y=(y_{1}y_{2}\cdots y_{t})$ in a permutation group $S_{n}$ are conjugate.
My idea was that two cycles $x=(x_{1}x_{2}\cdots x_{m})$ and $y = (y_{1}y_{2}\cdots y_{t})$ in $S_{n}$ are conjugate if and only if the cycles are of the same length (i.e., $m=t$) and $\exists f \in S_{n}$ such that $\forall x_{i}$, $f(y_{i}) = x_{i}$.
Here's my attempt at a proof:
$(\implies)$ Suppose $x = (x_{1} x_{2} \cdots x_{m})$ and $y = (y_{1}y_{2}\cdots y_{t})$ in $S_{n}$ are conjugate. Then $\exists f \in S_{n}$ such that $ f y f^{-1} = x \\ \implies \, f(y_{1} y_{2} \cdots y_{t})f^{-1}=(x_{1}x_{2}\cdots x_{m}) \\ \implies \, (f(y_{1})f(y_{2})\cdots f(y_{t})) = (x_{1},x_{2},\cdots, x_{m})$.
So, $f(y_{1})=x_{1}$, $f(y_{2})=x_{2}, \cdots , f(y_{t})=x_{m} \, \implies \, t = m$.
$(\Longleftarrow)$ Suppose $x=(x_{1}x_{2}\cdots x_{m})$ and $y = (y_{1}y_{2}\cdots y_{t})$ in $S_{n}$ are of the same length (i.e., $m = t$) and suppose also that $\exists f \in S_{n}$ such that $\forall i, f(y_{i}) = x_{i}$.
Then, we must have that $(f(y_{1})f(y_{2})\cdots f(y_{t})) = (x_{1}x_{2} \cdots x_{m}) \, \implies \, f(y_{1}y_{2}\cdots y_{t})f^{-1} = (x_{1}x_{2},\cdots, x_{m}) \, \implies \, x$ and $y$ are conjugate.
I doubt that this is true, but I made a bonafide attempt. If it's not true, could somebody please tell me what is a necessary and sufficient condition for two cycles to be conjugate and steer me in the direction of how to prove it?
I do know that in $S_{n}$, two permutations are conjugate iff they are of the same cycle type. And, in fact, that's next on my list of things I have to prove.
Thanks.