There is no ring map $\mathbb C[x] \to \mathbb C[x]$ swapping the prime ideals $(x-1)$ and $(x)$ I want to prove the following proposition: 

There is no ring map $\mathbb C[x] \to \mathbb C[x]$ such that the induced map between the spectra $\mathbb A^1_{\mathbb C}$ swaps the prime ideals $(x-1)$ and $(x)$ (sending one to the other) while keeping other prime ideals (i.e. $(x-a) \mapsto (x-a)$).

If not necessary, I don't want to assume this ring homomorphism $\mathbb C[x] \to \mathbb C[x]$ sends $1$ to $1$ and I also don't want to assume it is a vector space homomorphism over $\mathbb C$.
I believe there must be a general theorem or proposition setting restriction on the ring homomorphisms. So both slick proofs and general proofs are welcome in here! 
 A: Let $\varphi : \mathbb{C}[x] \to \mathbb{C}[y]$ be a ring homomorphism (which, as per the comments, I will assume sends $1$ to $1$; the zero homomorphism does not induce a map on spectra). The use of $x$ and $y$ is to make it easier to tell when we're talking about the source vs. the target.
$\varphi(\mathbb{C})$ must be a subfield of $\mathbb{C}[y]$, and since polynomials of positive degree are not invertible, it must be a subfield of $\mathbb{C}$. So $\varphi$ sends scalars to scalars, although the resulting map $f : \mathbb{C} \to \mathbb{C}$ may be complicated, e.g. it may be complex conjugation, any element of the very large automorphism group of $\mathbb{C}$, or it may even fail to be surjective. 
$\varphi$ is determined by its restriction to $\mathbb{C}$ and $\varphi(x) \in \mathbb{C}[y]$, which is some polynomial $g(y)$. If $a \in \mathbb{C}$, the maximal ideal $(y - a)$ in $\text{Spec } \mathbb{C}[y]$ is then sent by $\varphi^{\ast}$ (the action of $\varphi$ on spectra) to the kernel of the map
$$\mathbb{C}[x] \ni \sum a_i x^i \mapsto \sum f(a_i) g(a)^i \in \mathbb{C}.$$
Suppose first that $g(y)$ is constant. Then the kernels of all of these maps are the same, which contradicts the requirement that all maximal ideals other than $(x)$ and $(x - 1)$ are fixed. So we may assume that $g$ is nonconstant.
Next, suppose that $f$ is not surjective. Then $\mathbb{C}$ must be transcendental over $f(\mathbb{C})$: it cannot be algebraic since $f(\mathbb{C})$ is algebraically closed. Hence there is some $a$ such that $g(a)$ takes a value transcendental over $f(\mathbb{C})$: for this value of $a$ the kernel of the map above is $(0)$, which again contradicts the requirement that all maximal ideals other than $(x)$ and $(x - 1)$ are fixed. So we may assume that $f$ is surjective, hence an automorphism of $\mathbb{C}$.
Now we can describe the effect of $\varphi^{\ast}$ as follows: $g(a)$ has minimal polynomial $(x - g(a))$ over $\mathbb{C}$, so the kernel of the above map is $(x - f^{-1}(g(a))$. In other words, $\varphi^{\ast}$ induces the map
$$\mathbb{C} \ni a \mapsto f^{-1}(g(a)) \in \mathbb{C}$$
on maximal spectra. By hypothesis, we want this map to fix every $a \neq 0, 1$, hence we want
$$\forall a \neq 0, 1 : f(a) = g(a).$$
So $g(a)$, a polynomial, must agree with an automorphism of $\mathbb{C}$ for all inputs other than $0, 1$, so must satisfy the identities $g(x + y) = g(x) + g(y)$ and $g(xy) = g(x) g(y)$ as long as $x, y, x + y, xy \neq 0, 1$. These are all polynomial identities in two variables, and so if they're satisfied on a Zariski dense subset of $\mathbb{C}^2$ then they're always satisfied. The conclusion is that $g(a)$ must also be an automorphism of $\mathbb{C}$, and hence (e.g. by linearity) must agree with $f$ everywhere, so
$$\forall a : f(a) = g(a).$$
But the only polynomials which are linear are the polynomials $g(x) = px$, and among these the only one which preserves multiplication is the identity $g(x) = x$. So $f(x) = x$ also, and we conclude: 

The only ring homomorphism $\varphi : \mathbb{C}[x] \to \mathbb{C}[x]$ inducing a map on maximal spectra sending $a$ to $a$ for $a \neq 0, 1$ is the identity.

This would have been a lot less work if you had allowed $\mathbb{C}$-linearity, which allows the following more geometric argument: you want a map on spectra which agrees with the identity on a Zariski-dense subset of the spectrum, and so it must agree with the identity everywhere. This corresponds to having $f(x) = x$ from the start in the argument above. 
A: $\def\CC{\mathbb{C}}$Let $f:\CC[x]\to \CC[x]$ be a ring map such that $f^{-1}(x-2)\in (x-2)$, so that there is a $u\in \CC[x]$ with $$x-2=f(x-2)f(u).$$ If we had $f(x-2)\in \CC$, then the whole image of $f$ would be in $\CC$, because $f(\CC)\subseteq \CC$, and the above displayed equation would be impossible. It follows then from that equation, by looking at degrees, that there are $\alpha$ and $\beta$ in $\CC$ such that $f(x)=\alpha x+\beta$ and that $f(u)$ is a nonzero scalar. Looking even more closely, we see that $2\alpha+\beta-2=0$.
If we now also suppose that $f^{-1}(x-3)\in (x-3)$ we also have that $3\alpha+\beta-3=0$, and then $\alpha=1$ and $\beta=0$. 
These only two conditions already imply that $f(x)=x$, and then of course we do not have $f^{-1}(x)\in (x-1)$. 
