$G$ is semiregular implies its centralizer is transitive How do I prove that the centralizer of every semiregular group is transitive?  This is Exercise 4.5 in [Wielandt, Finite Permutation Groups].
Recall that a permutation group $G \le S^\Omega$ is semiregular if $G_\alpha=1, \forall \alpha \in \Omega$.  Let $Z:=C_{S^\Omega}(G)$.  We need to show $G$ is semiregular implies $Z$ is transitive.
Here's what I know. The text gives a proof of the converse (i.e. that $Z$ is transitive implies $G$ is semiregular). In addition, I can prove that $G$ is transitive implies $Z$ is semiregular, as follows. Fix $\alpha \in \Omega$, and let $z \in Z_\alpha$.  We need to show $z=1$. Let $\beta \in \Omega$. By transitivity of $G$, $\exists g \in G$ such that $\alpha^g=\beta$.  Then, $\beta^z = \beta^{g^{-1}zg}=\alpha^{zg}=\alpha^g=\beta$, whence if $z$ fixes $\alpha$ then it fixes all other points as well, i.e. $Z_\alpha=1$. 
 A: I will use the same notation that Wielandt uses.  Let $\alpha, \beta \in\Omega$.  I will create $x \in S^\Omega$ so that $\alpha^x=\beta$ and then show $x\in Z$.
Define $x$ based on the cases:
(1) If $\beta\in \alpha^G$, then set $\alpha^x := \beta$ and for $g\in G$, $(\alpha^g)^x := \beta^g$ for all $g\in G$ and $\gamma^x := \gamma$ for $\gamma \in \Omega - \alpha^G$.
(2) If $\beta \notin \alpha^G$, then set $\alpha^x:=\beta$ and for $g\in G$, $(\alpha^g)^x:=\beta^g$, $(\beta^g)^x:=\alpha^g$ and $\gamma^x := \gamma$ for $\gamma \in \Omega - \{\alpha,\beta\}^G$.
Note that if $\alpha^g=\alpha^h$, then $gh^{-1}\in G_\alpha$ and so $g=h$ as $G$ is semiregular.  This gives that $x$ is a well defined set map $x: \Omega \to \Omega$.
The orbits of $G$ on $\Omega$ partition $\Omega$ and I have defined $x$ on each each orbit.  It remains to see that $x$ is a bijection on $\Omega$.  In case (1), $x$ restricts to an injective set map $x: \alpha^G \to \alpha^G$ and is therefore a bijection.  In case (2), $x$ restricts to injective set maps $x:\alpha^G \to \beta^G$ and $x:\beta^G\to \alpha^G$.  Since $G$ is semiregular, $|\alpha^G| = |\beta^G|$ and is therefore a bijection.  In either case, $x\in S^\Omega$.
Note by construction, $x\in Z$.
