Centraliser in a group extension Given a group $N$ with a subgroup $D \leq N$, for an automorphism $x \in {\rm Aut}(N)$ such that $x$ normalises $D$ but no power of $x$ centralises $D$, before yesterday I would have thought that if we form the group $G = N \rtimes \langle x \rangle$ then $C_G(D)=C_N(D)$. 
The logic was that in $G$ it holds that $\langle x \rangle \cap C_N(D) = \{1\}$, and I thought that was enough.
However, I ran into an example in which this is evidently not true, and then I built a "toy" one:
$$G= (C_7 \rtimes C_3) \times C_3 = (\langle a\rangle \rtimes \langle b \rangle) \times \langle c\rangle$$
Where if we consider $\langle a\rangle \lhd \langle a, bc\rangle \lhd \langle a, bc, bc^2 \rangle = G$ then it is evident that both $bc$ and $bc^2$ do not centralise $\langle a \rangle$, and yet in the end group $C_G(\langle a\rangle) = C_3 = \langle c \rangle$.
So my question is: how do I check if the condition $C_G(D)=C_N(D)$ holds if, even when I have full knowledge of the automorphism $x$, I have no guarantee that this will imply equality of centralisers? Is there some criterion I am not aware of, or is there some additional condition for which something can be said? (e.g. $N$ simple).
 A: Let $\operatorname{Aut}_N(D)$ be the image of $N_N(D)/C_N(D)$ in ${\rm Aut}(D)$.
We claim that in the situation described above $C_G(D) = C_N(D)$ if and only if any nontrivial $x \in G/N$ defines an automorphism of $D$ that is not contained in $\operatorname{Aut}_N(D)$.
Without loss of generality, suppose that the order of $x$ is a prime number $p$.
Suppose that $C_G(D) = C_N(D)$. Then $N_N(D)/C_N(D) \lhd N_G(D)/C_G(D)$. Since $x$ normalises $D$, $N_N(D) \lhd N_G(D)$ with index $p$. Then any element in $N_G(D)$ whose coset is nontrivial in $N_G(D)/N_N(D) \cong \langle x \rangle$ defines an automorphism not contained in ${\rm Aut}_N(D)$.
Now suppose that $x$ defines an automorphism not contained in ${\rm Aut}_N(D)$. Then from the N/C theorem $N_G(D)/C_G(D) > N_N(D)/C_N(D)$, as every element in the latter embeds in $N_G(D)/C_G(D)$ (induced by the inclusion $N \to G$), but $N_G(D)/C_G(D)$ contains the automorphism induced by $x$ so the inclusion is strict. 
Then in particular $[N_G(D)/C_G(D) : N_N(D)/C_N(D)] = p$. Since $[N_G(D):N_N(D)]=p$, then $|C_G(D)| = |C_N(D)|$ which, together with the fact that $C_N(D) \lhd C_G(D)$, implies our claim.
