Can an isomorphism of subgroups of a finite group always be extended to an automorphism on the group?

The cyclic subgroups generated by $(12)$ and $(12)(34)$ in $S_4$ are isomorphic, but there is no automorphism of $S_4$ sending one to another: every automorphism of $S_4$ is inner and $(12)$ and $(12)(34)$ are not conjugate, being of different cycle types.

Example 1. A normal subgroup isomorphic to a nonnormal subgroup.
$H=\{(1),\ (1\ 2)(3\ 4),\ (1\ 3)(2\ 4),\ (1\ 4)(2\ 3)\}$ and $K=\{(1),\ (1\ 2),\ (3\ 4),\ (1\ 2)(3\ 4)\}$ are isomorphic subgroups of $S_4$, but $H$ is a normal subgroup of $S_4$ and $K$ is not. An automorphism of a group must take normal subgroups to normal subgroups.

Example 2. An Abelian example.
$G=\langle(1\ 2\ 3\ 4),\ (5\ 6)\rangle$ is an Abelian group of order $8$. The subgroups $H=\langle(1\ 2\ 3\ 4)^2\rangle$ and $K=\langle(5\ 6)\rangle$ are isomorphic subgroups of order $2$. No automorphism of $G$ maps $(1\ 2\ 3\ 4)^2$ to $(5\ 6)$, since the odd permutation $(5\ 6)$ is not a square; also, the quotient groups $G/H$ and $G/K$ are nonisomorphic.