Automorphism of $\Bbb Q[x]$ 
Question: Find a nonidentity automorphism $\varphi$ of $\Bbb Q[x]$ such that $\varphi^2$ is the identity automorphism of $\Bbb Q[x]$.

Is $\varphi(\Bbb Q[x]) = \Bbb Q[-x]$ a solution? I think it is correct by I have no confidence to write it down.

If it's confidence you need:
Let $\varphi$ such that $\varphi^2=id$.
Since it's a ring-homomorphism, $\varphi|_\mathbb{Q}$ must be a field-automorphism. But $\text{Aut}(\mathbb{Q})=\{id\}$. Therefore, it holds $\varphi(q(x))=q(\varphi(x))\ \forall q(x)\in\mathbb{Q}[x]$.
Moreover, $\varphi(x)=p(x)$ for some $p(x)\in\mathbb{Q}[x]$.
But, then, $\varphi^2(x)=\varphi(p(x))=p(\varphi(x))=p(p(x))$
Notice that $\deg(p(p(x)))=\deg^2(p(x))$. From $p(p(x))=x$, $\deg(p(x))=1$.
Hence $(\exists a\neq0\ \ p(x)=ax+b)\Rightarrow p(p(x))=a^2x+ab+b$
Hence, $a=-1\vee(a=1\wedge b=0)$. But the second case yeilds the identity, therefore $\varphi$ must send $x\mapsto b-x$ for some $b\in\mathbb{Q}$.
So, I strongly advise you to verify that $x\mapsto-x$ extends to a homomorphism, because it's the simplest candidate you can hope for.
