We know that, If there exists none other than only subgroup $H$ of order $m$, then $H$ is normal in $G$. When It comes to simple group, There should exist at the least $2$ subgroups of same order. To generate subgroups of The same order as $H$ consider $gHg^{-1}$ where $g$ is some element in $G\setminus H$.
My question is whether can we generate all the subgroups of the same order in the above manner? i.e. taken a subgroup $K$ of order $m$ is there an element $g$ in $G\setminus H$ such that $gHg^{-1}=K$? at least for the group being simple.
Thanks in advance
The answer is no, and there are many examples.
Consider for example the alternating group $G = A_6$. Let $H$ be a subgroup generated by $3$-cycle, and let $K$ be a subgroup generated by a product of two disjoint $3$-cycles. Then $H \cong K$, but $H$ and $K$ are not conjugate subgroups in $G$. You can easily generalize this to find many more examples.