Prove that in a group of permutation $S_7$ every two elements with order $10$ are conjugated. Prove that this fact is not truth in $S_9$.
In $S_7$ elements with order $10$ have a character: $$(a_1a_2)(b_1b_2b_3b_4b_5)$$because $10=5\cdot2$ and $2+5=7$.
If two elements $a,b$ are conjugated then we have: $$\exists _{g\in G} \text{ } ag=gb$$ Let's take $a=(a_1a_2)(b_1b_2b_3b_4b_5)$ so $|a|=10$ and some $g$ about which we know nothing so $$g=\begin{pmatrix} a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & a_7 \\ b_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 \end{pmatrix}$$ where $b_1 , b_2 , b_3 , b_4 , b_5 , b_6 , b_7 \in S_7 $ and $b_1 \neq b_2 \neq b_3 \neq b_4 \neq b_5 \neq b_6 \neq b_7 $.
Then: $$ag=(a_1a_2)(a_3a_4a_5a_6a_7)(a_1b_1)(a_2b_2)(a_3b_3)(a_4b_4)(a_5b_5)(a_6b_6)(a_7b_7)=(b_1a_2b_2a_1)(b_3a_4b_4a_5b_5a_6b_6a_7b_7a_3)$$
The next we would have to do the $gb$ action and prove that $ag = gb$.
This is probably possible, but I think this way is too much time-consuming and I don't see any facts here that would lead me to a smarter solution.
Have you got some idea to this solution?
First observe that for any permutation $\tau$ and a cycle $(a_1 \,\, a_2 \,\, \ldots a_n )$, $$\tau (a_1 \,\, a_2 \,\, \ldots a_n ) \tau ^{-1} =(\tau(a_1) \,\, \tau(a_2) \,\, \ldots \tau(a_n) ).$$ So conjugate of an $n-$cycle is also an $n-$cycle.
Every element of order $10$ in $S_7$ will have a $(5,2)$ cycle structure. So to show they are conjugates, we need to come up with a $\tau$ such that $$\tau (a_1 \,\, a_2)(b_1 \,\, b_2 \,\, b_3 \,\, b_4 \,\, b_5)\tau^{-1}=(c_1 \,\, c_2)(d_1 \,\, d_2 \,\, d_3 \,\, d_4 \,\, d_5).$$ Observe that \begin{align*} \tau (a_1 \,\, a_2)(b_1 \,\, b_2 \,\, b_3 \,\, b_4 \,\, b_5)\tau^{-1} & =\tau (a_1 \,\, a_2)\color{red}{\tau^{-1}}\,\, \color{red}{\tau}(b_1 \,\, b_2 \,\, b_3 \,\, b_4 \,\, b_5)\tau^{-1}\\ & = (\tau(a_1) \,\, \tau(a_2)) \,\, (\tau(b_1) \,\, \tau(b_2) \,\, \tau(b_3) \,\, \tau(b_4) \,\, \tau(b_5)). \end{align*} So any permutation $\tau$ that does the following will help us with conjugation: $$(\tau(a_1) \,\, \tau(a_2)) \,\, (\tau(b_1) \,\, \tau(b_2) \,\, \tau(b_3) \,\, \tau(b_4) \,\, \tau(b_5))=(c_1 \,\, c_2)(d_1 \,\, d_2 \,\, d_3 \,\, d_4 \,\, d_5).$$
Regarding $S_9$, the cycle structure of $(1 2 3 4 5)( 6 7)$ and $(1 2 3 4 5)( 6 7)(8 9)$ are different so they cannot be conjugates but both of them have order $10$ in $S_9$.