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

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

My answer is no. $$\operatorname{Aut} (\mathbb{Z)}$$ is isomorphic to $$Z_2$$ and $$\operatorname{Aut} (Z_3)$$ is also isomorphic to $$U(3)$$, which is isomorphic to $$Z_2$$. But $$\mathbb Z$$ is not isomorphic to $$Z_3.$$ Correct? Thanks

• Sorry I meant$Z_3$ – ramanujan Dec 17 '18 at 11:22
• @Derek Holt I edited it to$Z_3$. Am I correct now? – ramanujan Dec 17 '18 at 11:23
• Yes, you are correct. Also there's no need for introducing $U(3)$, $Aut(\mathbb{Z}_3)$ has exactly two elements, thus it has to be $\mathbb{Z}_2$. – freakish Dec 17 '18 at 11:24
• @freakish thanks – ramanujan Dec 17 '18 at 11:25

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: $$\operatorname{Aut}(Id)\cong Id\cong\operatorname{Aut(C_2)}$$
(These are the only two groups $$G$$ with $$\operatorname{Aut}(G)\cong Id$$. See here for a proof.)
You can even use non-isomorphic finite groups of the same order. The smallest example is $$Aut(C_4\times C_2)\cong Aut(D_8)\cong D_8$$.