Problem: Let $\alpha$ and $\beta$ be transcendental elements over $\mathbb{Q}$, show that there exists automorphism $\sigma$ of $\mathbb{C}$, such that $\sigma(\alpha) = \beta$
My attempt: Because $\alpha$ and $\beta$ are transcendental over $\mathbb{Q}$, then they are naturally isomorphic to $\mathbb{Q}(\alpha) \cong \mathbb{Q}(x)$, $\mathbb{Q}(\beta) \cong \mathbb{Q}(y)$. Here $\mathbb{Q}(x), \mathbb{Q}(y)$ are quotient field of polynominal rings over $\mathbb{Q}[x]$ and $\mathbb{Q}[y]$. Let $\phi: \mathbb{Q}[x] \rightarrow \mathbb{Q}[y]; \text{ satisfies } \phi(x) = y$. Then $\phi$ is obviously an isomorphism, and I can extend $\phi$ to $\psi: \mathbb{Q}(x) \rightarrow \mathbb{Q}(y)$. Therefore, $\mathbb{Q}(x) \cong \mathbb{Q}(y)$ (This is trivial tho) and so, I get $\mathbb{Q}(\alpha) \cong \mathbb{Q}(\beta)$. From the way how I defined $\psi$, it satisfies $\psi(\alpha) = \beta$. This is what I needed.
I thought I'm done now, but I also think I'm missing something(because this looks trivial and I did almost nothing). Is my solution fine?
(edit) My first attempt was not enough, so I try giving the 2nd one.
2nd attempt: Let $\phi$ be defined above. Now I extend $\phi$ to $\tilde{\phi}: \mathbb{C}[\alpha] \rightarrow \mathbb{C}[\beta]$, $\tilde{\phi}(\alpha) = \beta$. And in this case, because $\alpha$ and $\beta$ are in $\mathbb{C}$,$\mathbb{C}[\alpha]$ and $\mathbb{C}[\beta]$ are just $\mathbb{C}$. But I don't know this $\tilde{\phi}$ is an automorphism or not. $\tilde{\phi}$ satisfies $\tilde{\phi}(\alpha) = \beta$ and $\tilde{\phi}(z) = z$ for any $z \in \mathbb{C}-\{\alpha\}$. I think if I can say $\tilde{\phi}(\beta) = \alpha$, then I can say $\tilde{\phi} \in Aut(\mathbb{C})$, but nothing ensures $\tilde{\phi}$ holds that. Maybe I misunderstanding something.
If $\tilde{\phi}$ is an automorphism, then I'm done.