Here is an expanded version of my comment (it turns out you don't need the generalized Heisenberg group, just the standard one). My claim is that the theorem as stated is false.

Consider the Heisenberg group, $H$, defined over $\mathbb{Z}/2$. This finite group may be described simply as upper triangular matrices in $SL_3(\mathbb{Z}/2\mathbb{Z})$ with ones along the diagonal. Let $b$ be the elementary matrix $$ \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}. $$ As Johannes Hahn suggested, the group $A := C_{Aut(H)}(b)$ is defined to be all the automorphisms of $H$ that fix $b$. Given Marty Isaacs comment, $C_H(A)$ is defined by:

$$ C_H(A) := \{ g \in H : \forall a \in A, a(g) = g \}. $$

However, if $\phi : H \to H$ is any automorphism, then $\forall a, b \in H$, we have $\phi([a,b]) = [\phi(a), \phi(b)]$. Thus, if $a$ and $b$ commute, then their images commute. As $\phi$ is surjective it follows that $\phi(Z(H)) \leq Z(H)$. Since $Z(H) = \mathbb{Z}/2 \mathbb{Z}$ and since $\phi$ is injective: $\phi(Z(H)) = Z(H)$. Thus, $\phi$ must fix the element, call it $c$, that generates the center of $H$.

By definition of $A$, for any $a \in A$ we have $a(b) = b$. Further, by the previous paragraph, $\forall a \in A, a(c) = c$. Thus $C_H(A)$ contains $\left< b, c \right>$ which is precisely $\mathbb{Z}/2 \times \mathbb{Z}/2$. This is not cyclic, so the theorem stated in this question does not hold for all finite groups.

Final remarks. There seems to be something special about two here. Maybe the theorem is true if your finite group has order that is not divisible by two? Or perhaps you have a different definition of $C_G(A)$ in mind?

Here is an expanded version of my comment (it turns out you don't need the generalized Heisenberg group, just the standard one). My claim is that the theorem as stated is false.

Consider the Heisenberg group, $H$, defined over $\mathbb{Z}/2$. This finite group may be described simply as upper triangular matrices in $SL_3(\mathbb{Z}/2\mathbb{Z})$ with ones along the diagonal. Let $b$ be the elementary matrix $$ \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}. $$ As Johannes Hahn suggested, the group $A := C_{Aut(H)}(b)$ is defined to be all the automorphisms of $H$ that fix $b$. Given Marty Isaacs comment, $C_H(A)$ is defined by:

$$ C_H(A) := \{ g \in H : \forall a \in A, a(g) = g \}. $$

However, if $\phi : H \to H$ is any automorphism, then $\forall a, b \in H$, we have $\phi([a,b]) = [\phi(a), \phi(b)]$. Thus, if $a$ and $b$ commute, then their images commute. As $\phi$ is surjective it follows that $\phi(Z(H)) \leq Z(H)$. Since $Z(H) = \mathbb{Z}/2 \mathbb{Z}$ and since $\phi$ is injective: $\phi(Z(H)) = Z(H)$. Thus, $\phi$ must fix the element, call it $c$, that generates the center of $H$.

By definition of $A$, for any $a \in A$ we have $a(b) = b$. Further, by the previous paragraph, $\forall a \in A, a(c) = c$. Thus $C_H(A)$ contains $\left< b, c \right>$ which is precisely $\mathbb{Z}/2 \times \mathbb{Z}/2$. This is not cyclic, so the theorem stated in this question does not hold for all finite groups.

Final remarks. There seems to be something special about two here. Maybe the theorem is true if your finite group has order that is not divisible by two? Or perhaps you have a different definition of $C_G(A)$ in mind?

Update: Geoff Robinson's answer shows that there is nothing special about two here.

Here is an expanded version of my comment (it turns out you don't need the generalized Heisenberg group, just the standard one). My claim is that the theorem as stated is false.

Consider the Heisenberg group, $H$, defined over $\mathbb{Z}/2$. This finite group may be described simply as upper triangular matrices in $SL_2(\mathbb{Z}/2\mathbb{Z})$ with ones along the diagonal. Let $b$ be the elementary matrix $$ \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}. $$ As Johannes Hahn suggested, the group $A := C_{Aut(H)}(b)$ is defined to be all the automorphisms of $H$ that fix $b$. Given Marty Isaacs comment, $C_H(A)$ is defined by (the alternative interpretation of $C_H(A)$ is handled by Derek Holt's comment):

$$ C_H(A) := \{ g \in H : \forall a \in A, a(g) = g \}. $$

However, if $\phi : H \to H$ is any automorphism, then $\forall a, b \in H$, we have $\phi([a,b]) = [\phi(a), \phi(b)]$. Thus, if $a$ and $b$ commute, then their images commute. As $\phi$ is surjective it follows that $\phi(Z(H)) \leq Z(H)$. Since $Z(H) = \mathbb{Z}/2 \mathbb{Z}$ and since $\phi$ is injective: $\phi(Z(H)) = Z(H)$. Thus, $\phi$ must fix the element, call it $c$, that generates the center of $H$.

By definition of $A$, for any $a \in A$ we have $a(b) = b$. Further, by the previous paragraph, $\forall a \in A, a(c) = c$. Thus $C_H(A)$ contains $\left< b, c \right>$ which is precisely $\mathbb{Z}/2 \times \mathbb{Z}/2$. This is not cyclic, so the theorem stated in this question does not hold for all finite groups.

Final remarks. There seems to be something special about two here. Maybe the theorem is true if your finite group has order that is not divisible by two? Or perhaps you have a different definition of $C_G(A)$ in mind?

Here is an expanded version of my comment (it turns out you don't need the generalized Heisenberg group, just the standard one). My claim is that the theorem as stated is false.

Consider the Heisenberg group, $H$, defined over $\mathbb{Z}/2$. This finite group may be described simply as upper triangular matrices in $SL_3(\mathbb{Z}/2\mathbb{Z})$ with ones along the diagonal. Let $b$ be the elementary matrix $$ \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}. $$ As Johannes Hahn suggested, the group $A := C_{Aut(H)}(b)$ is defined to be all the automorphisms of $H$ that fix $b$. Given Marty Isaacs comment, $C_H(A)$ is defined by:

$$ C_H(A) := \{ g \in H : \forall a \in A, a(g) = g \}. $$

However, if $\phi : H \to H$ is any automorphism, then $\forall a, b \in H$, we have $\phi([a,b]) = [\phi(a), \phi(b)]$. Thus, if $a$ and $b$ commute, then their images commute. As $\phi$ is surjective it follows that $\phi(Z(H)) \leq Z(H)$. Since $Z(H) = \mathbb{Z}/2 \mathbb{Z}$ and since $\phi$ is injective: $\phi(Z(H)) = Z(H)$. Thus, $\phi$ must fix the element, call it $c$, that generates the center of $H$.

By definition of $A$, for any $a \in A$ we have $a(b) = b$. Further, by the previous paragraph, $\forall a \in A, a(c) = c$. Thus $C_H(A)$ contains $\left< b, c \right>$ which is precisely $\mathbb{Z}/2 \times \mathbb{Z}/2$. This is not cyclic, so the theorem stated in this question does not hold for all finite groups.

Final remarks. There seems to be something special about two here. Maybe the theorem is true if your finite group has order that is not divisible by two? Or perhaps you have a different definition of $C_G(A)$ in mind?