Consider an $n$-degree irreducible polynomial $f(x)$ on $\mathbb{Q}$. Generally, it has $n $ different roots in $\mathbb{C}$. Denote them as $\{ \alpha_1, \alpha_2, \ldots, \alpha_n \}$. Each of the roots can generate a simple extension of $\mathbb{Q}$, i.e., $\mathbb{Q}(\alpha_i )$ with $1\leq i \leq n $.
The problem is, what is the relation between these $n $ field extensions?
By the way, an explicit isomorphism $\mathbb Q(\alpha_i) \to \mathbb Q(\alpha_j)$ is the morphism that fixes $\mathbb Q$ and sends $\alpha_i \mapsto \alpha_j$.
To prove that this is an isomorphism, it's probably best to show that $\mathbb Q[x]/(f(x))$ is isomorphic to each $\mathbb Q(\alpha_i)$ via the morphism that fixes $\mathbb Q$ and sends $x \mapsto \alpha_i$.
Consider the homomorphism of rings, $$ \phi_i : \mathbb Q[x] \to \mathbb Q[\alpha_i] \\ \phi_i(g(x)) = g(\alpha_i).$$ Here, $\mathbb Q[\alpha_i]$ is the ring generated by $\alpha_i$ over $\mathbb Q$. (We'll soon prove that $\mathbb Q[\alpha_i] = \mathbb Q(\alpha_i)$; this will be a nice bonus result!)
Since $f(x)$ is irreducible, it is the minimal polynomial of $\alpha_i$ (up to multiplication by a constant factor), so the kernel of $\phi_i$ is the principal ideal $(f(x))$. The image of $\phi_i$ is the whole of $\mathbb Q[\alpha_i]$. So by the first isomorphism theorem, $\phi_i$ descends to an isomorphism of rings, $$ \mathbb Q[x]/(f(x)) \cong \mathbb Q[\alpha_i].$$ This isomorphism sends the equivalence class of $x$ to $\alpha_i$.
But since $f$ is irreducible and $\mathbb Q[x]$ is a PID, the quotient $\mathbb Q[x]/(f(x))$ is in fact a field. Hence $\mathbb Q[\alpha_i]$ is equal to $\mathbb Q(\alpha_i)$, the field generated by $\alpha_i$.
Composing these isomorphisms, you learn that $\mathbb Q(\alpha_i)$ is isomorphic to $\mathbb Q(\alpha_j)$, via the morphism fixing $\mathbb Q$ and sending $\alpha_i \mapsto \alpha_j$.
