If the centralizer of an element of a group is abelian, then can two elements inside it be conjugates? I have a small doubt, which might be trivial too. But I am unable to see it. Let $p$ and $q$ be odd primes such that  $q \ | \ p-1.$ Consider the group $G = A \rtimes \langle b \rangle$, where $A$ is a finite $p$-group and $b$ is of order $q.$ If we have the group ring $FG$ over the field $F$ with p elements, then its unit group $\mathcal{U}(FG)$ turns out to be of the form $(1+I) \rtimes \mathcal{U}(F \langle b \rangle),$ where I is an ideal of $FG$ and $1+I$ is a p-group. Here the centralizer of $b$ in $1+I, C_{1+I}(b),$ turns out to be an abelian group. Then is it possible that for two elements $z_i, z_j \in C_{1+I}(b),$ there exists some $u \in \mathcal{U}(FG)$ such that $u^{-1}z_iu = z_j?$
 A: We have $G = P \rtimes Q$ (I have renamed your $A$ as $P$) with $P$ a finite $p$-group, and $Q = \langle b \rangle$ cyclic of order $q$ with $q|p-1$, and we are assuming that $C := C_P(b)$ is abelian. I will attempt to prove that it is impossible for distinct elements $x,y \in C$ to be conjugate in $G$, and we will work by induction on $P$. I am afraid that my proof is somewhat lengthy!
Since $b$ centralizes $x$ and $y$, if they are conjugate then they must be conjugate by an element $g \in P$. So clearly $P$ cannot be abelian.
Now $b$ normalizes the characteristic subgroup $\Omega_1(Z(P))$ of $P$ (i.e. the elements of order $1$ or $p$ in $Z(P)$). Since $q|p-1$, the irreducible modules for $Q$ in characteristic $p$ have dimension $1$, so $Q$ normalizes some subgroup $\langle z \rangle$ of $\Omega_1(Z(P))$ of order $p$. Let $Z = \langle z \rangle$.
We will use the general result that, if a finite group $Q$ of order coprime to the prime $p$ acts on a $p$-group $P$, then $P = C_P(Q)[P,Q]$.
Let $D/Z := C_{P/Z}(b)$. Then applying the above result to the action of $Q$ on $D$, we have $[Q,D] \le Z$, so $D=CZ$. In particular, $D/Z$ is abelian, and by inductive hypothesis applied to $P/Z \rtimes Q$, two distinct elements of $D/Z$ cannot be conjugate, and so we must have $xZ = yZ$. We can assume that $y=xz$ (which implies that $C=D$).
Now let $X = C_P(x)$. Since $g^{-1}xg = y = xz$, and $g \not\in X$, but the image $gZ$ of $g$ in $P/Z$ lies in the subgroup $Y$ with $Y/Z = C_P(xZ)$. Since $x$ has at most $p$ conjugates in $Y$, we must have $|Y:X| = p$, and $Y = \langle X,g \rangle$. Furthermore, since $b$ centralizes $x$, it normalizes both $X$ and $Y$.
We have $b^{-1}gb = g^kw$ for some $k$ with $1 \le k \le p-1$ and $w \in X$. Then $g^{-1}xg=xz$ implies that $(g^kw)^{-1}x(g^kw) = g^{-k}xg^k = xz$. But $g^{-k}xg^k = xz^k$, so $k=1$.
So $b$ centralizes $Y/X$, but then by the result above applied to the action of $Q$ on $Y$, we have $[Q,Y] \le X$ and $Y = C_Y(Q)[Q,Y]$, so some element $gv$ with $v \in X$ must lie in $C_P(b)$. But $C_P(b)$ is abelian, so $gv$ and hence also $g$ must centralize $x$, contradiction.
