Conjugate subgroups are normal? Suppose $A \subset C$ and $B \subset C$. Assume $A$ and $B$ are conjugate subgroups, that is $cAc^{-1}=B$ for some $c \in C$. Is the following statement true?
$A=B$ if and only if $A$ and $B$ are normal subgroups of $C$
Question: If we assume $A$ and $B$ are conjugate subgroups of $C$, are they normal subgroups? 
The statement above is a statement that I come out on my own to aid me to understand relationship between regular covering and conjugacy class.
EDIT: How to relate conjugacy class with regular covering/ normal covering? 
 A: Fixing that one error hopefully means that you are well on your way to understanding the subtleties (conjugation, normalcy) here. It is very much possible that $cAc^{-1}=A$ (or $A=B$ if you wish) for some but not all $c\in C$. The latter is equivalent to $A$ being normal in $C$, $A\unlhd C$.
The in-between cases can be described using what's called the normalizer of $A$ in $C$. This is the set
$$
N_C(A)=\{c\in C\mid cAc^{-1}=A\}.
$$
It is a standard application of the subgroup criterion to show that $N_C(A)$ is a subgroup of $C$. Leaving that and the following facts as exercises:


*

*We always have $A\unlhd N_C(A)\le C$.

*$A$ is a normal subgroup of $C$ if and only if $N_C(A)=C$.

*The normalizer $N_C(A)$ is the largest subgroup $H\le C$ with the property $A\unlhd H$.


As an example let us look at the case $C=S_4$, and $A=\langle(12)\rangle$ - the cyclic subgroup of order generated by the 2-cycle $(12)$. Here it is not too hard to show that (depends on how familiar you with the algebra of permutations)
$$
N_C(A)=\{1_C, (12), (34), (12)(34)\}.
$$
For example with $c=(34)$ you get $cAc^{-1}=A$ because $(12)$ and $(34)$ commute. In this case $N_C(A)$ is properly between $A$ and $C$. When we pick $c$ outside $N_C(A)$, say $c=(23)$, we get
$$
B=cAc^{-1}=\langle(13)\rangle\neq A.
$$
