Prove a property about the centralisator Let G be a group and $U \subseteq G$ a subgroup. Let $x \in G$ be arbitrary.
How to show that $C_G(xUx^{-1})=xC_G(U)x^{-1}$
where $C_G(U):=\{g\in G : gu=ug$ $\forall u\in U\}$
For the first direction I have:
$z \in C_G(xUx^{-1}) \Rightarrow z(xux^{-1})=(xux^{-1})z$ for all $u\in U$ $(xux^{-1} \in xUx^{-1})$
We have: $z=x(ux^{-1}zxu^{-1})x^{-1}$ We now have to show that $ux^{-1}zxu^{-1} \in C_G(U)$ ??
 A: A more abstract line: For any $\sigma \in \text{Aut}(G)$, one can (easily) show that $\sigma(C(U)) = C(\sigma(U))$. The result then follows by taking $\sigma$ to be conjugation by $x$.
A: Suppose $z\in C_G(xUx^{-1})$.
We want to show that $z\in xC_G(U)x^{-1}$, i.e., that $z=xyx^{-1}$ for some $y\in C_G(U)$, i.e., that $y=x^{-1}zx\in C_G(U)$.
Thus, we have, for arbitrary $u\in U$:
$yu=x^{-1}zxu=x^{-1}zxux^{-1}x=x^{-1}xux^{-1}zx=ux^{-1}zx=uy$,
so it looks like y is in that centralizer.
Does that explain the forward direction well enough?
A: Your way is correct, so far.
To complete it, we show that $u$$x^{-1}zxu^{-1}$ is in $C_G(U)$
Let $u_0 \in U$, we conclude:
$$ux^{-1}zxu^{-1}u_0=u_0ux^{-1}zxu^{-1} $$
$\iff$ 
$$ux^{-1}zxu^{-1}x^{-1}xu_0=u_0x^{-1}xux^{-1}zxu^{-1}$$
$\iff$ 
$$ux^{-1}xu^{-1}x^{-1}zxu_0=u_0x^{-1}zxux^{-1}xu^{-1} $$
$\iff$ 
$$x^{-1}zxu_0=u_0x^{-1}zx $$
$\iff$ 
$$zxu_0x^{-1}=xu_0x^{-1}z $$
Which is correct because $z \in C_G(xUx^{-1})$
A: the crucial fact is that the centralizer is defined entirely in terms of constraints of the form
$$
ha = ah
$$
which are trivially preserved under any automorphism. in particular if $h^x = xhx^{-1}$ then we have:
$$
h^xa^x = a^xh^x
$$
this is to say that if $h$ commutes with $a$ iff $\forall Ax\in G$ $xhx^{-1}$ commutes with $xax^{-1}$
