$\operatorname{Aut} (G)$ is isomorphic to $\operatorname{Aut} (H)$ then is it necessary that $G$ is isomorphic to $H$?

Besides your example, there is even an example with finite groups, as $$ {\rm Aut}(S_3)\cong {\rm Aut}(C_2\times C_2)\cong S_3, $$ but $S_3$ is of course not isomorphic to $C_2\times C_2$.

As another example, both the trivial group $Id$ and the cyclic group of order two $C_2$ have trivial automorphism group: $$\operatorname{Aut}(Id)\cong Id\cong\operatorname{Aut(C_2)}$$

This is the smallest possible example...

(These are the only two groups $G$ with $\operatorname{Aut}(G)\cong Id$. See here for a proof.)