Recently I was studying automorphism of groups, and I found that ${\rm Aut}(\mathbb{Q})$ is isomorphic to $\mathbb{Q}^\ast$. I have proved it. I also showed if two groups are isomorphic then so are their automorphism groups.
After this I moved on to check if ${\rm Aut}(\mathbb{Q}[\sqrt{2}])\simeq \mathbb{Q}^\ast$ as groups. For this purpose, I was curious to show if $\mathbb{Q}\simeq \mathbb{Q}[\sqrt{2}]$ is true but got stuck. How can I show this?
You can't show any of your last two statements, both are false.
First $\mathbb{Q}$ is not isomorphic to $\mathbb{Q}[\sqrt{2}]$ as groups. That's because given two $\mathbb{Q}$-linear spaces $A$, $B$ these are isomorphic as groups if and only if they are isomorphic as $\mathbb{Q}$-linear spaces. And $\mathbb{Q}$ has dimension $1$ over $\mathbb{Q}$ while $\mathbb{Q}[\sqrt{2}]$ is of dimension $2$.
Secondly $\rm Aut(\mathbb{Q}[\sqrt{2}])$ is not isomorphic to $\mathbb{Q}^*$. Indeed, $\mathbb{Q}^*$ has exactly two elements of finite order, namely $1$ and $-1$. While $\rm Aut(\mathbb{Q}[\sqrt{2}])$ has more, at least three: $$f_1(x)=x$$ $$f_2(x)=-x$$ $$f_3(a+b\sqrt{2})=b+a\sqrt{2}$$