Let $\mathbb{L}/\mathbb{K}$ be a finite normal extension with Galois group $G = \operatorname{Gal}(\mathbb{L}/\mathbb{K})$. Let $\mathbb{L}^H := \left\lbrace~ x \in \mathbb{L} \mid \sigma(x) = x \quad \forall \sigma \in H ~\right\rbrace$ denote the fixed field of $H$.
Let $H, K$ be two subgroups of $G$. Does $H \cong K$ imply $\mathbb{L}^H \cong \mathbb{L}^K$?
I tried going through an example:
Let $\mathbb{L} := \mathbb{Q}(\sqrt[3]{7}, \zeta_3)$ (where $\zeta_3$ is the third root of unity) be the splitting field of $x^3 - 7 \in \mathbb{Q}[x]$. Clearly, the Galois group is $S_3$. Take two subgroups of $S_3$ in the isomorphism class of $S_2$; $H_1 := \big\lbrace~\operatorname{Id}, (12)~\big\rbrace$ and $H_2 := \big\lbrace~\operatorname{Id}, (23)~\big\rbrace$. Now, $\mathbb{L}^{H_1} = \mathbb{Q}(\sqrt[3]{7}\zeta_3^2) \cong \mathbb{Q}(\sqrt[3]{7}) = \mathbb{L}^{H_2}$. So it seems to work for this case; though I'm not sure if I just got lucky.