Let $G$ be a finite group. Suppose $H$ and $K$ are two isomorphic maximal subgroups of $G$, then can we claim that there must be an automorphism $\alpha\in {\rm Aut}(G)$ such that $\alpha(H)=K$?
If $H$ and $K$ don’t have to be maximal subgroups, then I’ve already found a counterexample of order $8$. In $G:=C_2\times C_4$, there are three subgroups isomorphic to $C_2$ and only one of them is characteristic in $G$. Therefore no automorphism of $G$ can map the characteristic one to the others.
I failed to find a counterexample in the case where $H$ and $K$ are maximal subgroups of $G$. Could you give me some ideas? Any help is appreciated. Thank you!
There are many such examples, I'm afraid. One source of examples is the maximal subgroups of classical groups. The smallest example I could find were two classes of $\mathrm{PSL}_2(7)$ in $\mathrm{PSp}_6(q)$, for $q\equiv \pm 7\bmod 16$, $q\neq 7$.
More generally, almost every irreducible character of a simple group $G$ yields a maximal embedding of $G$ into some classical group. (This isn't quite right, but it's good enough for these purposes.) The multiplicities of irreducible character degrees of finite groups tend to infinity as the group order tends to infinity, so there are unboundedly many such classes in general.