the conjugates of r-cycles are r-cycles. Let $0<r \leq n$ be positive integers, and consider an $r$-cycle $\sigma(j_1\ j_2...j_r) \in S_n.$ I want to show that $\tau \sigma \tau^{-1}=(\tau(j_1)\tau(j_2)...\tau(j_3))$.
Let $\tau \in S_n.$ We can write $\sigma$ as $\sigma=(j_1j_r)(j_1j_{r-1})...(j_1j_2)$, so \begin{align*}
         \tau \sigma \tau^{-1}&=\tau \left((j_1j_r)(j_1j_{r-1})...(j_1j_2)\right)\tau^{-1}\\
         &= \tau (j_1j_r)\tau^{-1}\ \tau (j_1j_{r-1})\tau^{-1}...\tau (j_1j_2)\tau^{-1}
     \end{align*} 
    The last equality is true since for any permutation $\tau \in S_n$, $\tau^{-1}$ will take any number $j_t$ to $j_1$ with $0<t\leq r.$ 
I dont know if this is true or should I prove that in another way?
I would appreciate any help or hints with that, thanks in advance.

You don't need to split it into a product of transpositions. It is much easier. If $1\leq k<r$ then where does $\tau\sigma\tau^{-1}$ send $\tau(j_k)$? 
$\tau\sigma\tau^{-1}(\tau(j_k))=\tau\sigma(j_k)=\tau(j_{k+1})$
And similarly $\tau\sigma\tau^{-1}(\tau(j_r))=\tau(j_1)$. 
