Let $\overline{\mathbb{Q}}$ be an algebraic closure of $\mathbb{Q}$. Then, does there exists finite extension $E_1,E_2$ of $\mathbb{Q}$ inside $\overline{\mathbb{Q}}$, such that $E_1\neq E_2$ but $E_2\cong E_2$?
Here, extensions $E_1,E_2$ of $\mathbb{Q}$ are isomorphic if there is a bijective map from $E_1$ to $E_2$ which preserves addition and multiplication in the fields.
Instead of $\mathbb{Q}$, we may take any field, but there we get no such examples (line $\mathbb{R}$, $\mathbb{C}$).
For example, for $\mathbb{R}$, we can take $\overline{\mathbb{R}}=\mathbb{C}$. Now between $\mathbb{R}$ and $\overline{\mathbb{R}}$, there are no other fields. Hence we do not get $E_1,E_2$ such that $\mathbb{R}\subseteq E_1 ,E_2 \subseteq \overline{\mathbb{R}}$ but $E_1\neq E_2$.
