Are the only single element normal subgroups precisely the elements that make up the center of a group? 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 $G$.
 A: 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.
A: The only issue, as mentioned by others, is that the only single element subgroup is the trivial group. But it's a common exercise along these lines to show normal subgroups of order $2$ are in the center for the reasons you say. 
A: The only single element (normal) subgroup of a group is $\{e\} $, where  $e $ is the identity element. No other single element subgroup is possible.  
A: If $g$ is the unique element of order $n$, then $n=1$ or $n=2$ and $g$ must be central. Perhaps this is what you mean, Leo?
A: As others have pointed out, every subgroup must contain the identity, so there is only one single-element subgroup, namely $\{1_G\}$.
However, I think there is a way to rescue this claim (and its proof) in a non-trivial way.  Define a normal subset to be a subset $S\subseteq G$ with the property that $gS = Sg$ for all $g\in G$.  Note, this is not standard terminology; however, it does work well with the standard phrase "normal subgroup", in that a normal subgroup would just be a subgroup that is a normal subset in this sense.
What you have actually proved is that the only single-element normal subsets are precisely the elements that make up the center of a group.
