How do we know that automorphisms on polynomials have a polynomial like form? Let $F$ be a field and $\sigma:F[x]\to F[x]$ be automorphism, $\sigma(a) = a$ for all $a\in F$. I'm supposed to show that $\sigma(f(x)) = f(ax+b)$ for some $a\not = 0$ and $b$ in $F$.
Now I've got a solution that my professor gave me that seems to assume that the automorphism must have the form $\sigma(f(x)) = f(p(x))$ for some $p(x)\in F[x]$. So my question is how does $\sigma$ being an automorphism on $F[x]$ and $\sigma(a) = a$ for all $a\in F$ give us that $\sigma(f(x) = f(p(x))$ for some $p(x)\in F[x]$, why can't there be some weirder looking automorphism?
I've looked at Automorphisms of $F[x]$, however the only solution seems to make the same assumption that my professor makes.
 A: Since $\sigma$ fixes $F$, that is,
$\sigma(a) = a, \; a \in F, \tag 1$
we have, for any
$r(x) = \displaystyle \sum_0^n r_i x^i \in F[x], \tag 2$
$\sigma(r(x)) = \sigma \left ( \displaystyle \sum_0^n r_i x^i \right ) = \displaystyle \sum_0^n \sigma(r_i) \sigma(x^i) = \sum_0^n r_i (\sigma(x))^i; \tag 3$
thus, $\sigma$ is determined by its value on $x$, $\sigma(x)$; since by definition,
$\sigma:F[x] \to F[x], \tag 4$
we must have
$\sigma(x) = p(x) \in F[x]; \tag 5$
thus, from (3), we see that
$\sigma(r(x)) = r(\sigma(x)) = r(p(x)). \tag 6$
A: Let $\sigma(x) = p(x)$.
If $f(x) = \sum_{i=0}^n a_i x^i$, then using the fact that $\sigma$ is a ring homomorphism,
$$\sigma(f(x)) = \sigma \left( \sum_{i=0}^n a_i x^i \right) = \sum_{i=0}^n \sigma(a_i) \sigma(x)^i = \sum_{i=0}^n a_i p(x)^i = f(p(x)).$$
More generally, if $A$ is any $F$-algebra, then every $F$-algebra homomorphism $F[x] \to A$ takes the form $f(x) \mapsto f(a)$ for some $a \in A$.
A: Hint:
If $\sigma \in$ Aut (F[$x$]) such that $\sigma$ (a)=a, for all $\in$ F, deg$\sigma(f(x))$= deg$f(x)$ , for all $f(x)\in$ F[$(x)$].
In particular $\sigma(x)= ax + b$, where $a \neq0$.
