Let $\text{Homeo}(\mathbb{R})$ denote the group of self-homemorphisms of $\mathbb{R}$. If $G$ is a finite group, is $G$ isomorphic to a subgroup of $\text{Homeo}(\mathbb{R})$?
(Edit: this was previous denoted $\mathrm{Aut}(\mathbb{R})$, so in the comments, $\mathrm{Aut}(\mathbb{R})$ has to be interpreted as $\text{Homeo}(\mathbb{R})$.)
I am just posting my comment above as an answer.
There is a normal subgroup of $\text{Homeo}(\mathbb{R})$ of index 2, namely the group $\text{Homeo}^{+}(\mathbb{R})$ of order-preserving homeomorphisms. Thus, if $G$ is a subgroup of $\text{Homeo}(\mathbb{R})$, then $G\cap\text{Homeo}^{+}(\mathbb{R})$ is a normal subgroup of $G$ of index 1 or 2. For a finite subgroup of $\text{Homeo}^{+}(\mathbb{R})$, the orbits of the induced action on $\mathbb{R}$ must just be singleton sets (just the least element in each finite orbit). Thus, the only finite groups isomorphic to a subgroup of $\text{Homeo}(\mathbb{R})$ are $\{e\}$ and a cyclic group of order 2.