Let $D$ be a skew field, assumed to be finite dimensional over its centre $Z(D)$. Let $\sigma\in\mathrm{Aut} (D)$ be an automorphism of $D$, and let $D^{\sigma}$ be the set of elements in $D$ fixed by $\sigma$: this is also a skew field.
It is clear that $Z(D)^\sigma\subseteq Z(D^\sigma)$. (Indeed, if $x$ is central in $D$, then it also commutes with every element in $D^\sigma$, so if $x\in Z(D)^\sigma$, then $x\in Z(D^\sigma)$.) Does the converse hold, that is, is $Z(D)^\sigma= Z(D^\sigma)$?
To put it another way: does the Schur index of $D$ always divide that of $D^\sigma$?
Consider the case of Hamiltonian quaternions $D=\Bbb{H}$. Let $\sigma$ be conjugation by the quaternion $i$. Then $D^{\sigma}=\Bbb{R}(i)=\Bbb{C}$, and therefore $Z(D^{\sigma})=D^{\sigma}$. But $\Bbb{R}=Z(D)=Z(D)^{\sigma}$, so the inclusion is strict in this case.
This example generalizes as follows. Let $F$ be a number field, and $K$ a maximal subfield of $D$. We know that $K=F(\rho)$ for some element $\rho\in K$. If $\sigma$ is conjugation by $\rho$, we see that $K= D^{\sigma}$.