Does there exist any $x\in S_{10}$ such that $x(1\ \ 2\ \ 3)x=(9\ \ 10)$?

Question

I have tried it a lot by many trial and error methods, I didn't get any such $x\in S_{10}$. One this that I can observe that $\operatorname{Order}(x)>2$, because if $\operatorname{Order}(x)=2$ then $x(1\ \ 2\ \ 3)x=x(1\ \ 2\ \ 3)x^{-1}=(x(1)\ \ x(2)\ \ x(3))\ne(9\ \ 10)$
But I even can't prove that there doesn't exist such $x$.
I haven't any idea how to solve the problem. Can anybody give a solution to it? Thanks for assistance idn advance.

Answer 1

$x(1\,2\,3)x$ must be an even permutation. But $(9\,10)$ isn't.

In general, to solve $xax=b$ in $S_n$where $a$ and $b$ are given permutations, note that this is the same as $(xa)^2=ba$, so this is soluble iff $ba$ is a square in $S_n$.