Let $$\sigma = \begin{pmatrix} 1&2&3&4&5&6\\ 4&5&2&1&6&3\end{pmatrix}\in S_6.$$ Then $\sigma = (1\text{ }4)(2\text{ }5\text{ }6\text{ }3)$ and thus $\sigma^{-1}=(4\text{ }1)(3\text{ }6\text{ }5\text{ }2)$. Then ${\rm sgn}(\sigma)={\rm sgn}(\sigma^{-1})=1\Rightarrow \sigma,\sigma^{-1}\in A_6$. I now have to find $\tau \in A_6$ such that $\tau \sigma \tau^{-1}=\sigma^{-1}$.
I know that $$\tau=\begin{pmatrix}1&2&3&4&5&6\\ 4&3&2&1&6&5\end{pmatrix}$$ solves it since $$\tau \sigma \tau^{-1}=(\tau(1)\tau(4))(\tau(2)\tau(5)\tau(6)\tau(3))=(4\text{ }1)(3\text{ }6\text{ }5\text{ }2)=\sigma^{-1},$$but $\tau \not \in A_6$ and this is where I'm stuck.
Am I going in the right direction?
Any help is appreciated. Thank you for your time.