If $x \in G$, is it possible that $C = \{g^{-1}xg : g \in G \}$ is a subgroup of $G$? Can $C$ be a normal subgroup of $G$? (What are necessary and sufficient conditions to be such a subgroup?)
Showing that $C$ is a subgroup involves, I think, finding a way to restrict $x$ and/or $g$ so that you can get closure to hold, however I'm not sure how to do that (an intersection perhaps). Also, I think $C$ is normal already, the problem is just verifying the subgroup criterion. Here is the normal proof: Let $g_1 \in G$ be arbitrary. Then, $g_1 C g_1^{-1} = g_1 \{g^{-1}xg : g \in G \} g_1^{-1} = g_1g^{-1}xgg_1^{-1}$such that $g \in G$. So, $g_1g^{-1}xgg_1^{-1} = (gg_1^{-1})^{-1}x(gg_1^{-1}) \in C$ since $gg_1^{-1} \in G$. So, $g_1Cg_1^{-1} \subseteq C$. Clearly, $C \subseteq g_1Cg_1^{-1}$. Let me know your ideas please!
You are correct that, if $C$ is a subgroup, then it is a normal subgroup. However, $C$ is only a subgroup when $x$ is the identity element; to see this quickly, one may note that the relation defined by: $$x\sim y \Longleftrightarrow \exists g\in G [x=gyg^{-1}]$$ is an equivalence relation, dividing the group into conjugacy classes. So, $C$ happens to be the conjugacy class of $x$. The trouble is that all subgroups must contain the identity $e$ and $g^{-1}eg = g^{-1}g = e$ - so the identity is conjugate only to itself. Therefore, if $C$ is to be a subgroup, it must be the conjugacy class of $e$ and must equal the trivial subgroup $\{e\}$.
(That said, in any case, the subgroup generated by $C$ is normal - and is the normal closure of $\{x\}$. There's not much you can say about the structure of this subgroup in general, however)