When does a permutation of the roots define an automorphism?

Answering the modified version of the question.

The splitting field $K$ of $f(x)=x^3-10$ over $F=\Bbb{Q}(\sqrt{-3})$ is the cubic extension $$ K=F(\root3\of{10}). $$ This is because the primitive third root of unity $\omega=e^{2\pi/3}=(-1+\sqrt{-3})/2\in F$, and the zeros of $f(x)$ are $x_k=\omega^k\root3\of{10}$ with $k=0,1,2$.

In this case we do get a cyclic group $C_3$ as the Galois group $Gal(K/F)$. The reason why we cannot use all the permutations of $\{x_0,x_1,x_2\}$ is that $\omega\in F$ must be fixed by any $F$-automorphism. And $\omega$ is the ratio of two consequtive roots. So if $\sigma\in Gal(K/F)$ then $$ \sigma(x_1)=\sigma(\omega x_0)=\sigma(\omega)\sigma(x_0)=\omega\sigma(x_0) $$ and similarly $$ \sigma(x_2)=\sigma(\omega x_1)=\omega \sigma(x_1)=\omega^2\sigma(x_0). $$ This shows that $\sigma(x_1)$ and $\sigma(x_2)$ are fully determined once we know $\sigma(x_0)$. The permutations in this Galois group must observe this constraint, and thus only three permutations work.

A possibly more interesting case of the same phenomenon is present in Lord Shark's example polynomial $x^3+x^2-2x-1$. There you can show that if $r$ is one of the roots, $r^2-2$ is another. Such a hidden polynomial relation between those two roots must be observed by all the automorphisms, and thus not all permutations will be allowed.

The Galois group of $X^3-10$ over $\Bbb Q$ is $S_3$, so in fact every permutation of its roots does define an automorphism. A better example is $X^3+X^2-2X-1$ where the Galois group is indeed cyclic of order $3$.

In short the permutations of the roots that induce automorphisms are those coming from the elements of the Galois group, and that's about really all one can say.

You don't need to know anything about permuting roots to show that $i \mapsto -i$ defines an automorphism of $\mathbb{Q}(i)$. The point is that $\mathbb{Q}(i) \cong \mathbb{Q}[x]/(x^2 + 1)$, so a homomorphism from $\mathbb{Q}[x]/(x^2 + 1)$ to another field of characteristic $0$ (or any other $\mathbb{Q}$-algebra) is uniquely determined by where it sends $x$, which is subject only to the constraint that $x^2 = -1$.

So there are exactly two homomorphisms $\mathbb{Q}(i) \to \mathbb{Q}(i)$ because $\mathbb{Q}(i)$ has exactly two elements, namely $i$ and $-i$, which square to $-1$. I'm not picking a permutation of the roots when I do this, just choosing where a single root goes, namely $i$, which uniquely determines where $-i$ goes.

Here's a more complicated but similar example. Consider now $\mathbb{Q}(z)$ where $z^5 = 1$ is a primitive fifth root of unity, which is a degree $4$ extension. This is equivalently $\mathbb{Q}[x]/(x^4 + x^3 + x^2 + x + 1)$ (this requires a little proof) which shows that homomorphisms $\mathbb{Q}(z) \to \mathbb{Q}(z)$ are uniquely determined by where they send $z$, which is subject to the constraint that $z^5 = 1$ but $z \neq 1$. Thus there are four such homomorphisms, sending $z$ to $z, z^2, z^3, z^4$, and we find that the Galois group is

$$(\mathbb{Z}/5\mathbb{Z})^{\times} \cong \mathbb{Z}/4\mathbb{Z}.$$

In particular, because all the other fifth roots of unity are powers of $z$, a homomorphism is completely determined by where it sends $z$, and once that's known you have no choice about where the other roots go.