Expanding an isomorphism between subgroups of $G$ to an automorphism on $G$

No: e.g., if $H_1$ is a subgroup of the centre of $G$ and $H_2$ is not, then there can be no automorphism of $G$ mapping $H_1$ to $H_2$.

Thanks to Gerry Myerson and SteamyRoot for supplying specific examples.

The conjecture also fails in abelian groups. E.g., take $G = \Bbb{Z}_2 \times \Bbb{Z}_4$ and take $H_1$ and $H_2$ to be the subgroups generated by $x = (0, 2)$ and $y = (1, 0)$ respectively. Then the isomorphism $\phi : H_1 \to H_2$ that maps $x$ to $y$ cannot be extended to $G$ because $x$ is divisible by $2$ but $y$ is not.

It's actually kind of hard to get even an automorphism of a subgroup that extends to an automorphism of the entire group (and when that is possible, there are usually many ways to do the extension part). There are some pretty strong conditions that are necessary for what you're asking about (isomorphism of subgroups extending to automorphism of whole group) to even be possible. First, there has to be an automorphism of $G$ which maps the subgroup $H_1$ to $H_2$, which requires that the subgroup lattice 'looks the same' (up to isomorphism types) from the perspective of either subgroup (giving a technical description of what 'looks the same' actually is, is a bit difficult and tedious, but if you look at enough reasonably small subgroup lattices, you should be able to intuit what all has to happen). Moreover, the 'looks the same' condition carries over to the subgroups of $H_1$ and their corresponding subgroups for $H_2$ (which limits which isomorphisms between them will work when it even is possible to do this).