Galois group acts transitively on the roots

Let $$F$$ be a field and $$p(x)$$ be an irreducible polynomial in $$F[x]$$. Let $$K$$ be a splitting field of $$f(x)$$.
Let $$R:=\{\alpha_1,\alpha_2, \alpha_3,\dots,\alpha_k\}$$ be the set of all roots of $$p(x)$$ in $$K$$. Then, it is easy to check that \begin{align*} \varphi_1:F(\alpha_1) &\to F(\alpha_2)\subseteq K \\ \alpha_1 & \to \alpha_2 \end{align*} is an $$F$$-isomorphism i.e. bijective ring homomorphism with $$\varphi_1(x)=x$$ for all $$x\in F$$.

Now, I want to extend the map $$\varphi_1$$ to an $$F$$-homomorphism \begin{align*} \varphi_2: F(\alpha_1)(\alpha_2)\to K \end{align*} Clearly, if such an extension $$\varphi_2$$ exists then it is determined by $$\varphi_2(\alpha_2)$$, and $$\varphi_2(\alpha_2)$$ must belong to $$R$$.

Question. How do we guarantee that such an extension $$\varphi_2$$ exist? Thanks!

Let $$h(t) \in F(\alpha_1)[t]$$ be the minimal polynomial of $$\alpha_2$$ over $$F(\alpha_1)$$, apply $$\varphi_1$$ to the coefficients of $$h$$ to obtain $$h^{\varphi_1}(t) \in \varphi_1(F(\alpha_1))[t]$$, set $$\varphi_2(\alpha_2) = \beta_2$$ for any root $$\beta_2 \in K$$ of $$h^{\varphi_1}(t)$$.
$$\beta_2$$ always exists because $$K/F$$ is normal. Continuing this way you'll have $$\varphi_k \in Aut(K/F)$$.