When does a normal subgroup contain precisely one non-identity conjugacy class? Every normal subgroup $N$ of a group $G$ is a union of conjugacy classes. Since every subgroup contains the identity, and the identity is in a class by itself, every normal subgroup already contains the conjugacy class of the identity.
So when is a normal subgroup comprised of exactly two conjugacy classes?
$N = \{1\} \cup \mathcal K$ 
Here is what I see so far:


*

*Unless $|N|=2$ and $N \leq Z(G)$, the subgroup must have trivial intersection with the center, since each element in the center is contained in its own conjugacy class.

*Since $|\mathcal K|$ is the index of the centralizer $C_G(k)$ of any $k\in\mathcal K$, and $|G:C_G(k)| = |N|-1$ divides $|G|$, we must have that G is divisible by the product $|N|(|N|-1)$ of two consecutive numbers. This also implies $|G|$ is even.


Any inner automorphism fixes $N$, but I don't know about outer automorphisms, so $N$ may not have to be a characteristic subgroup.
What is the full characterization of these types of normal subgroups? Do they have any important properties?
Edit: Ted is correct.
 A: All groups are finite.  The following is an old result of Wielandt, I believe.
Proposition: Such a subgroup $N$ must be an elementary abelian $p$-group, and every elementary abelian $p$-group is such a subgroup in a certain group $G$.
Proof: If two elements of $G$ are conjugate, then they have the same order. Hence every non-identity element of $N$ has the same order. The order cannot be composite since $g^a$ has order $b$ if $g$ has order $ab$. Hence $N$ is a $p$-group. The commutator subgroup of $N$ is characteristic in $N$ and so normal in $G$. Hence it is either all of $N$ or just $1$; however, in a non-identity $p$-group the commutator subgroup is always a proper subgroup. In particular, $N$ is abelian and every element has order $p$.
Now suppose such an elementary abelian group $N$ is given. Let $G=\operatorname{AGL}(1,p^n)$ be the set of affine transformations of the one-dimensional vector space $K$ over the field $K$ of $p^n$ elements. That is $G$ consists of all $\{ f : K \to K :x \mapsto \alpha x + \beta ~\mid~ \alpha,\beta \in K, \alpha \neq 0 \}$. Then every non-identity element of $N=K_+=\{ f : K \to K : x \mapsto x + \beta ~\mid ~ \beta \in K \}$ is conjugate under $K^\times = \{ f : K \to K : x \mapsto \alpha x ~\mid~ \alpha \in K, \alpha \neq 0 \}$. Thus the proof is complete. $\square$.
A: Your point (1) cannot be correct, as follows from
$$G=A_4\,\,,\,N:=\{(1)\,,\,(12)(34)\,,\,(13)(24)\,,\,(14)(23)\}$$
Here, $\,N\lhd G\,$ , indeed $\,N\cap Z(G)=1\,$ (which is not big deal as the center of this alternating group is trivial), but $\,N\,$ is the Sylow $\,2-\,$ subgroup of $\,A_4\,$ .
Number 2 is, imo, correct.
