If $G$ and $H$ are nonisomorphic group with same order then can we say that $Aut(G)$ is not isomorphic to $Aut(H)$?

We know that nonisomorphic groups may have isomorphic automorphism groups. As an example, you can think klein four group and $S_3$ since their automorphism group is isomorphic to $S_3$.

Now,I wonder If $G$ and $H$ are nonisomorphic group with same order then can we say that $\operatorname{Aut}(G)$ is not isomorphic to $\operatorname{Aut}(H)$ or can we find two nonisomorphic groups with same order and their
automorphism groups are isomorphic?

Answer

No.

You can check that Automorphism group of both Dihedral group($D_8$) and Direct product of $Z_2$ and $Z_4$ is Dihedral group($D_8$).

So we have two non-isomorphic groups with order $8$ and their Automorphism groups are the same group.

This is the smallest example of such groups.

http://groupprops.subwiki.org/wiki/Endomorphism_structure_of_direct_product_of_Z4_and_Z2

http://www.weddslist.com/groups/misc/autd8.html