For which non-constant rational functions $f(x)$ in $\mathbb{Q}(x)$ is there $\alpha$, algebraic over $\mathbb{Q}$, such that $\alpha$ and $f(\alpha) \neq \alpha$ are algebraic conjugates? More generally, can one describe the set of such $\alpha$ (empty/non-empty, finite/infinite etc.) if one is given $f$?
Examples:
- $f(x)=-x$: These are precisely the square roots of algebraic numbers $\beta$ such that there is no square root of $\beta$ in $\mathbb{Q}(\beta)$. There are infinitely many $\alpha$ and even infinitely many of degree $2$.
- $f(x)=x^2$: These are precisely the roots of unity of odd order, since $H(\alpha)=H(\alpha^2)=H(\alpha)^2$ implies $H(\alpha)=1$, so $\alpha$ a root of unity. Here $H(\alpha)$ is the absolute multiplicative Weil height of $\alpha$. There are infinitely many $\alpha$, but only finitely many of degree $\leq D$ for any $D$.
- $f(x)=x+1$: There is no $\alpha$. If there was and $P(x)$ was its minimal polynomial, then $P(x+1)$ would be another irreducible polynomial, vanishing at $\alpha$, with the same leading coefficient and hence $P(x+1)=P(x)$. Looking at the coefficients of the second highest power of $x$ now leads to a contradiction. Analogously for $f(x)=x+a$ if $a$ is any non-zero rational number.
So the existence of such an $\alpha$ and the set of all possible $\alpha$ seem to depend rather intricately on $f(x)$, which seems interesting to me. As I found no discussion of this question in the literature, I post it here.
UPDATE: Firstly thanks to all who have contributed so far! As Eric Wofsey pointed out, any solution $\alpha$ will satisfy $f^n(\alpha)=\alpha$ for some $n>1$, where $f^n$ is the $n$-th iterate of $f$. So one should consider solutions of the equation $f^n(x)-x=0$ or $f^p(x)-x=0$ for $p$ prime.
If the degree of $f$ is at least 2, one can always find irrational such solutions $\alpha$ with $f(\alpha) \neq \alpha$ by the answer of Joe Silverman. However, for his proof to work, we'd need to know that $f^k(\alpha)$ and $\alpha$ are conjugate for some $k$ with $0 < k <p$. I'm not enough of an expert to follow through with his hints for proving this, but if someone does, I'd be very happy about an answer!
If the degree of $f$ is 1, then $f$ is a Möbius transformation and all $f^n$ will have the same fixed points as $f$ (so there's no solution) unless $f$ is of finite order. In that case, if $f(x) \neq x$, the order is 2, 3, 4 or 6 (see http://home.wlu.edu/~dresdeng/papers/nine.pdf). By the same reference, in the latter three cases, $f$ is conjugate to $\frac{-1}{x+1}$, $\frac{x-1}{x+1}$ or $\frac{2x-1}{x+1}$, so it suffices to consider these $f$, which give rise to the minimal polynomials $x^3-nx^2-(n+3)x-1$ (closely related to the polynomial in GNiklasch's answer), $x^4+nx^3-6x^2-nx+1$ and $x^6-2nx^5+5(n-3)x^4+20x^3-5nx^2+2(n-3)x+1$, if my calculations are correct. If the order is 2, the map is of the form $\frac{-x+a}{bx+1}$ or $\frac{a}{x}$, which leads to $x^2+(bn-a)x+n$ or $x^2+nx+a$ respectively. So this case is somewhat degenerate, which explains the unusual behavior of $f(x)=-x$ and $f(x)=x+1$ above.
Best Answer
A first observation is that the field $\mathbb{Q}(\alpha)$ must admit a non-identity automorphism $\sigma$, since applying a rational function to $\alpha$ will not take us outside this field. Given such $\alpha$ and $\sigma$, you can immediately read off an $f$ by expressing $\sigma(\alpha)$ as a $\mathbb{Q}$-linear combination of $1, \alpha, \alpha^2,\ldots$, and there'll be additional non-polynomial (rational) expressions in terms of $\alpha$. This leads to plenty of examples, but now we want to turn this around and start from a given $f$.
Next, the application of $f$ can be iterated. If $g$ is the minimal polynomial of $\alpha$ over the rationals, $g \circ f$ will be a rational function whose numerator vanishes at $\alpha$, hence is divisible in $\mathbb{Q}[x]$ by $g$, hence vanishes at $f(\alpha)$ as well, so $g(f(f(\alpha))) = (g\circ f)(f(\alpha))=0$, so $f(f(\alpha))$ is another conjugate of $\alpha$, and so forth.
Since there are only finitely many conjugates to choose from, some iterate $f^{\circ n}(\alpha)$ must return to $\alpha$: in other words, $\alpha$ will be a root of (the numerator of) $f^{\circ n}(x)-x$ for some integer $n > 1$. (This leads to another argument ruling out $f(x)=x+1$ and friends, as has been noted in the comments.)
So in many cases you'll be able, given $f$, to find the $\alpha$ by picking an $n$ and factoring the numerator of $f^{\circ n}(x)-x$ into irreducible polynomials.
However, the $n$-th iterate of $f$ might be the identity function $x\in \mathbb{Q}(x)$ already, and the numerator to be factored might thus be zero! This happens in your first example $f(x)=-x$ with $n=2$, or with $$f(x)=1/x$$ again with $n=2$ which leads to reciprocal minimal polynomials $g(x)=x^{\mathrm{deg}(g)}g(1/x)$, or with (e.g.) $$f(x)=1-1/x$$ with $n=3$ which gives rise (among other things) to what D. Shanks in 1974 called The Simplest Cubic Fields (Math.Comp. 28, 1137-1152), with $g(x)=x^3+ax^2-(a+3)x+1$. Here, you can proceed by prescribing a degree for $g$, spelling out the condition that $g$ should divide the numerator of $g\circ f$, and comparing coefficients.