Let $E/F$ be a field extension of $F$, and let $H \leq \text{Aut}(E/F)$ be a subgroup of the automorphisms of $E$ fixing $F$. If $H$ is finite, then it is well-known that $H = \text{Aut}(E/E^H)$ and so $E^{H} = E^{\text{Aut}(E/E^H)}$, where $E^H$ is the fixed field of $H$.
Therefore, to find an example where equality does not hold, we necessarily require $H$ to be infinite. So I tried looking at $E = \mathbb{R}(x)$ and $F = \mathbb{R}$, which gives \begin{align} \text{Aut}(\mathbb{R}(x)/\mathbb{R}) \cong GL_2(\mathbb{R})\bigg/\begin{pmatrix}\lambda & 0 \\ 0 & \lambda\end{pmatrix}, \end{align} with $\lambda \in \mathbb{R}^\times$. An automorphism can be constructed by considering $g = \begin{pmatrix}a & b \\ c & d\end{pmatrix} \in GL_2(\mathbb{R})$ and defining a map $\phi: GL_2(\mathbb{R}) \to \text{Aut}(\mathbb{R}(x)/\mathbb{R}) $ by $g \mapsto \phi(g)$ where, for $f(x) \in \mathbb{R}(x)$, we have \begin{align} \phi(g)(f) = f\left(\frac{ax + c}{bx + d}\right). \end{align} The only problem is that I don't know how to find the fixed field. Say if I took $H = SO_2(\mathbb{R})\bigg/\begin{pmatrix}\lambda & 0 \\ 0 & \lambda\end{pmatrix} < \text{Aut}(\mathbb{R}(x)/\mathbb{R})$, then, to obtain the fixed field $\mathbb{R}(x)^H$, I would require all rational functions $f$ satisfying \begin{align} f\left(\frac{x\cos\theta + \sin\theta}{-x\sin\theta + \cos\theta}\right) = f(x). \end{align} How should I proceed from here, or, is there an easier example I can examine?