Complement of a normal subgroup is single conjugacy class The dihedral group of order $2n$ with $n$ odd has property that the complement of cyclic subgroup of order $n$ is a single conjugacy class. 

Q. Are there any other solvable groups in which complement of a normal subgroup is a single conjugacy class?


There are solvable groups in which every non-trivial coset of a fixed normal subgroup is conjugacy class; but my consideration is for single conjugacy class outside a normal subgroup.
 A: Let $N$ be any finite solvable group equipped with an action of a cyclic group $C=\{1,a\}$ of order $2$ which fixes no elements besides the identity.  For instance, $N$ could be any abelian group of odd order (not necessarily cyclic) with $a$ acting by negation.  Let $G=N\rtimes C$ be the semidirect product given by this action.  Then $G$ is solvable, $N$ is a normal subgroup, and I claim that $G\setminus N$ is a single conjugacy class.  Indeed, note that the element $(1,a)\in G$ does not commute with $(x,1)$ for any $x\neq 1$ in $N$ (since $a$ acts nontrivially on every non-identity element of $N$).  It follows that the centralizer of $(1,a)$ is just $\{(1,a),(1,1)\}$, and so the conjugacy class of $(1,a)$ contains $|G|/2$ elements.  But this conjugacy class is a subset of $G\setminus N$ which also has $|G|/2$ elements, so the conjugacy class must be all of $G\setminus N$.

Conversely, all finite examples are of this form: suppose $G$ is a finite group with a normal subgroup $N$ such that $G\setminus N$ is a conjugacy class.  Picking any $a\in G\setminus N$, we then have $|G\setminus N|=|G|/|C_G(a)|$.  In particular, since $|C(x)|\geq 2$, we have $|G\setminus N|\leq |G|/2$ and so $|N|\geq|G|/2$.  This can only happen if $|N|=|G|/2$, so $N$ must have index $2$ and $C_G(a)$ must have order $2$.
This means that $a$ must be an element of $G$ of order $2$ which commutes with no elements of $G$ except itself and $1$.  Moreover, we see that $G$ is the semidirect product of the subgroup $C$ generated by $a$ and $N$, and $a$ must act nontrivially on every non-identity element of $N$ since $a$ commutes with nothing except itself and $1$.
