For an algebraic number $\gamma$ which is not a root of unity, Baker's theorem gives
a bound (uniform in $n$) of the form $|\gamma^n - 1| > n^{-C}$ for some constant $C$
(only depending on $\gamma$).
In particular, if $s_n:=\prod |\alpha^n_i - 1|$, then Baker's method gives the following estimate uniform in $n$ (for $\alpha_i$ not a root of unity):
$$n \log \mathcal{M}(\alpha) + A \ge \log|s_n| \ge n \log \mathcal{M}(\alpha) - A - B \log|n|,$$
where $\mathcal{M}(\alpha)$ is the Mahler measure of $\alpha$. Proof: for $|\alpha_i| > 1$,
one has $\log|\alpha^n_i - 1| \sim n \log|\alpha_i|$ up to $O(1)$, which
gives rise to the term $n \log \mathcal{M}(\alpha)$;
the term $\log|\alpha^n_i - 1|$ for $|\alpha_i| \le 1$ is trivial to bound from above
and can be bound from below by Baker's Theorem. The logarithm of the Mahler measure is
the sum of $\log|\alpha_i|$ for the roots $|\alpha_i|> 1$. By a theorem of Kronecker this sum is positive if $\alpha$ is not a root of unity.
OTOH, if $\Phi_n(x)$ is the $n$th cyclotomic polynomial and
$t_n:=\prod |\Phi_n(\alpha_i)|$, then we deduce that
$\sum_{d|n} \log(t_n) = \log(s_n)$, and hence
$$\log(t_n) = \sum_{d|n} \log(s_{n/d}) \mu(d) \ge \varphi(n) \log \mathcal{M}(\alpha) -
d(n)(A + B \log(n)),$$
where $d(n)$ is the number of divisors of $n$. The bounds $\phi(n) \gg n^{1-\epsilon}$ and
$d(n) = n^{\epsilon}$ easily give the asymptotic relation
$$\log(t_n) \sim \varphi(n) \log \mathcal{M}(\alpha) \gg 1$$
as $n$ goes to infinity.
FWIW, the bounds of Baker (and the other bounds used above) are effective, so for any
particular $\alpha$ one could in principle find all $n$ with $t_n = 1$.
(n.b. Gelfond's estimate would give $\epsilon \cdot n$ instead of $B \log|n|$ as an error term,
which is enough to show that $s_n \rightarrow \infty$ but not $t_n$.)
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.
Best Answer
Assume that the roots $\zeta$ and $\xi^2/\zeta$ of your equation are roots of unity. Then by Vieta Theorem $\zeta+\xi^2/\zeta=-(\xi^2-\xi+1)$; equivalently sum of 5 roots of unity $\zeta+\xi^2/\zeta+\xi^2-\xi+1$ vanishes. All such relations are classified in Theorem 6 of paper by Conway and Jones. This implies the result.