If $G$ is a finite group of order $n$, why is it isomorphic to its centralizer in $S_n$? If $G$ is a finite group of order $n$, why is it isomorphic to its centralizer in $S_n$? Here, we embed $G$ in $S_n$ via the left regular representation.
From thinking a bit about the classification of covering spaces in algebraic topology, I suspect that this must be true, and I expect that I can probably prove it using topology if I wanted to.
Is there a representation/group-theoretic way of proving this?
Also, is the converse true? Ie, if $G$ is a subgroup of $S_n$ that is isomorphic to its own centralizer, then it must be transitive and have order $n$?
 A: I learned about the centralizer of the regular representation in M. Hall's Theory of Groups, Theorem 6.3.1 on page 86. In particular a very simple group theoretic proof is given.
Representation theory indicates that this should be true, since a permutation group is basically a module, and the endomorphism ring (also known as the centralizer) of a regular module is the opposite ring. Since a group is isomorphic to its opposite group via $g \mapsto g^{-1}$, it is no surprise that the centralizer of the left regular permutation representation is the right regular permutation representation.
The converse fails: a self-centralizing group need not move all of the points (for instance the cyclic group generated by an $n$ cycle in $S_{n+1}$ is self-centralizing as long as $n>1$). Even if a self-centralizing group moves all the points, it need not be transitive (for instance, the non-standard Klein 4 group, $\{ (), (12), (34), (12)(34) \}$).
However, if one assumes that the group is transitive, then an exercise in Wielandt's excellent textbook Permutation Groups on page 9 shows that a transitive group must be regular if it has the same order as its centralizer. The exercise has one show that the centralizer of a regular group is semiregular, and its order is equal to the number of fixed points in the original group's point stabilizer. For the orders to be equal, the point stabilizer must be the identity subgroup, that is, the group must be regular.
A: Suppose that $f: G \rightarrow G$ is a bijection that commutes with every left translation. Then $f(xy) = x f(y)$ for all $x, y \in G$. In particular choosing $y = 1$ gives $f(x) = xf(1)$ for all $x \in G$. Thus $f$ is a right translation by $f(1)$. Also, any right translation commutes with a left translation, so the centralizer of the left regular representation in $S_n$ is the right regular representation. Similarly, the centralizer of the right regular representation in $S_n$ is the left regular representation.
