Proof or counterexample: Galois conjugates of $\theta$ lie in $\mathbb{Z}[\theta]$ Let $f(x) \in \mathbb{Z}[x]$ be (edit: monic) irreducible such that if $\theta$ is a root of $f$ then $\mathbb{Q}(\theta)/\mathbb{Q}$ is a Galois extension. In other words, all Galois conjugates of $\theta$ lie in $\mathbb{Q}(\theta)$. Is it necessarily the case that all Galois conjugates of $\theta$ lie in $\mathbb{Z}[\theta]$?
Unfortunately such polynomials $f$ are fairly hard to come by in the wild, so 'fair' examples seem to be scarce. Quadratics are silly, I've seen a few contrived cubics, but none yield a counterexample, cyclotomics are silly etc.
Of course, we need an example so that $\mathbb{Z}[\theta]$ is not the full ring of integers...
...or the result may  be true...
I wouldn't be surprised if the proof/counterexample is trivial by the way...
EDIT
The original question has been answered below. Here are one or two follow up questions that I am still very interested in:


*

*(Suggested by JyrkiLahtonen): Suppose we insist that $\mathbb{Z}[\theta]$  is maximal among the subrings of the form $\mathbb{Z}[a]$ with $a$ an algebraic integer of $\mathbb{Q}(\theta)$; can the result be proven, or is there a counterexample? 

*What can we say about the denominators that appear? I wouldn't be surprised if the only primes that appear in denominators are those that divide the conductor of $\mathbb{Z}[\theta]$ (or $\mathbb{Z}[a]$) in the ring of integers... 

 A: I think there is the following rather trivial way of producing cubic counterexamples, namely using integer multiples of non-counterexamples.
Let $u=2\cos(2\pi/9)$. It is well known that the minimal polynomial of $u$
is $g(x)=x^3-3x-1$, and that $\sigma(u)=u^2-2=2\cos(4\pi/9)$ is a conjugate (the third conjugate is then $\sigma^2(u)=2\cos(8\pi/9)=2-u-u^2$ as the sum of all three conjugates is obviously zero). Therefore $L=\Bbb{Q}(u)$ is the splitting field of $g(x)$, and hence cyclic Galois of degree three.
Consider the number $\theta=3u$. It immediately follows that $L=\Bbb{Q}(\theta)$.
One of its conjugates is $$\theta':=\sigma(\theta)=3(u^2-2)=3u^2-6=\frac{\theta^2}3-6.$$
Because $\theta$ is an algebraic integer its (monic) minimal polynomial $m(x)$ has integer coefficients. So if $p(x)\in\Bbb{Z}[x]$ is any polynomial, then $p(x)=q(x)m(x)+r(x)$ for some at most quadratic $r(x)\in\Bbb{Z}[x]$. So if $p(\theta)=\theta'$ then also $r(\theta)=\theta'$. This is impossible because the representation of any element of $L$ as a quadratic polynomial in $\theta$ with rational coefficients is unique.
