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.
 A: $\renewcommand{\phi}{\varphi}\newcommand{\Q}{\mathbb{Q}}$First of all, any such $\phi$ will send each rational number (that is, the constants) to itself. This is because $\Q$ is the prime field here, i.e. the subfield generated by $1$.
Now the universal property of polynomial rings tells you that any ring homomorphism $\phi : \Q[x] \to \Q[x]$ which sends each rational to itself is described by assigning an arbitrary value to the image $z = \phi(x)$ of $x$. If $z$ is a constant, then clearly the image of $\phi$ is only $\Q$. If $z$ has degree higher than $1$, then the image $\Q[z]$ of $\phi$ won't contain polynomials of degree $1$.
Then $z = a x + b$, with $a \ne 0$. You want
$$
x = \phi(\phi(x)) = \phi(a x + b) = a (a x + b) + b = a^{2} x + b (a + 1).
$$
Excluding the identity ($a = 1$, and thus $b = 0$), all the remaining cases work, that is $a = -1$, $b$ arbitrary.
A: 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.
