So, a single element normal subgroup $n$ of group $G$ would be defined as $ \forall g \in G: gng^{-1} = n$. All elements of an abelian group would qualify, and so would more generally the elements of the center of a group, since $gng^{-1} = gg^{-1}n = n$ for $n\in \text{Z}(G)$.
Thus, to prove that these are the only elements for which this is true, we consider $gng^{-1} = n$. Right multiplying by $g$ yields $gng^{-1}g = ng$, which simplifies by associativity to $gn = ng$, implying that $g$ and $n$ commute. Thus, $n\in \text{Z}(G)$.
Is this line of reasoning correct?
Edit: My terminology was confusing in calling things "single element normal subgroup". They are not subgroups. I think a better question would have been to prove that the only normal subgroups $N$ where $\forall n \in N, \forall g \in G: gng^{-1} = n$ are subgroups of the center of $G$.
Best Answer
The phrase "single element normal subgroup" is confusing. The only subgroup with a single element is the subgroup consisting only of the identity.
Your argument correctly proves that the center of the group is the set of elements fixed by every inner automorphism.