Proving that $f(x)$ is irreducible in $F[x]$ iff $\phi(f(x))$ is. This is what I'm proving:

Let $F$ be a field. Let $\phi : F[x]\to F[x]$ be an isomorphism such that $\phi(a)=a$ for every $a\in F$. Prove that $f(x)$ is irreducible in $F[x]$ iff $\phi(f(x))$ is.

Here is my proof:

We prove the contrapositive, i.e. $\phi(f(x))$ is reducible iff $f(x)$ is.
($\Leftarrow$) Let $f(x)$ be reducible, i.e. $f(x)=g(x)h(x)$ for some $g(x),h(x)\in F[x]$ such that $0<\deg(g(x))<\deg(f(x))$ and $0<\deg(h(x))<\deg(f(x))$. Now since $\phi$ is an isomorphism we have $\phi(f(x))=\phi(g(x)h(x))=\phi(g(x))\phi(h(x))$ and thus $\phi(f(x))$ is reducible.
($\Rightarrow$) Let $\phi(f(x))$ be reducible in $F[x]$.
Now $\phi(f(x))=m(x)n(x)$ for some $m(x),n(x)\in F[x]$ such that $0<\deg(m(x))<\deg(\phi(f(x)))$ and $0<\deg(n(x))<\deg(\phi(f(x)))$. Now since $\phi$ is an isomorphism or rings, so is $\phi^{-1}$, and thus we have $f(x)=\phi^{-1} (\phi(f(x)))=\phi^{-1}(m(x)n(x))=$ $\phi^{-1}(m(x))\phi^{-1}(n(x))$, and hence $f(x)$ is reducible.

Can this be right since I never had to use the fact that $\phi(a)=a$ for all $a\in F$? If my proof is correct and I used this fact implicitly, I would like to know where. Thanks!
 A: user1952009 has a very good answer, but I have different one throwing some more light on the topic hopefully.
The point is, it's certainly true that $\phi(f(x))=\phi(g(x))\phi(h(x))$. However,$\phi(f(x))$ is reducible if and only if both $\phi(g(x))$ and $\phi(h(x))$ are not scalars. What if one of them were a scalar? That's why you need the condition that the function fixes $F$.
By the way, suppose that $\sigma$ is an automorphism from $F[x]$ to $F[x]$ fixing $F$.For $\sigma$ to be invertible,think about what the image of the element $x \in F[x]$ should be. Suppose it is some $y$, then $\phi(f(x))$ can be written as $f(y)$, with $y$ being the image of $x$. When will $f(y)$ cover the whole of $F[x]$? Only when $y$ itself is a one-degree polynomial, say $y=ax+b$, because only then can you define the inverse function, $\sigma^{-1}$ neatly, as $\sigma^{-1}(y)=\frac{y-b}{a}=x$, and $\sigma^{-1}$ fixes $F$. So now you know the general form of any such automorphism: it is of the form $f(ax+b)$, where $a,b$ are fixed scalars in $F$.
Now the proof is quite easy: If $f(x)=g(x)h(x)$, then $\phi(f(x))=\phi(g(x))\phi(h(x)) = g(ax+b)h(ax+b)$ and the degree of $g(a+bx)$ and $h(a+bx)$ are the same as $g(x)$ and $h(x)$ respectively, whence $\phi(f(x))$ is also reducible. Similarly, you can show the other way using the inverse function.
As a reply to user1952009's question, we can take any polynomial in $\mathbb{C}$ and send it to it's conjugate polynomial (conjugate each coefficient). For example, send $(1+i)x$ to $(1-i)x$. This is an automorphism of $\mathbb{C}[x]$ that does not fix $\mathbb{C}$.
A: You don't need the fact that $\phi$ fixes $F$.  If $A$ is a commutative ring with identity, an element $f \in A$ is defined to be irreducible if and only if whenever $g, h \in A$ and $f = gh$, you have $g$ or $h$ is a unit in $A$. If $\psi: A \rightarrow B$ is an isomorphism of rings, then it sends units to units, so it must send irreducible things to irreducibles.
Check that a polynomial $f \in F[X]$ is irreducible in this sense if and only if it is irreducible in the sense you're saying (use the fact that the units in $F[X]$ are exactly the nonzero constant polynomials).
Edit: By the way, the possibility that $\phi(g)$ is a scalar when $g$ is not, can't happen.  This is because $\phi$ sends units to units, so it sends elements of $F \setminus \{0\}$ to themselves.  In other words, $\phi$ may not be the identity on $F$, but the restriction of $\phi$ to $F$ is an isomorphism of $F$ onto itself.
More generally, if $f(X) \in F[X]$ is a polynomial of degree $d$, then $\phi(f)$ will also have degree $d$, regardess of whether or not $\phi$ fixes $F$.  
Proof: Since $\phi$ is completely determined by its effect on $F$ and on $X$, it is enough to show that $\phi(X) = aX+b$ for some $b,0 \neq a \in F$.  Already you know that $\phi(X)$ must have degree at least one.  But then its degree must be exactly one: every element of $F[X]$ is obtained by taking finitely many nonnegative powers of $X$, multiplying each power by some element of $F$, and adding everything together.  Hence the same must be true for $\phi(X)$.  So if $\phi(X)$ had degree $>1$, you would not be able to obtain linear polynomials from these combinations of $\phi(X)$. $\blacksquare$
