conjugacy classes in representation theory I have a question on conjugacy classes in this post, especially to this sentence: 
"if $g$ is a rotation of order $5$, then it is $K$-conjugate to each of $g^{13},g^{13^2},g^{13^3},g^{13^4}=g$". 
Here $g$ is the rotation in the group $D_5$ and $K=Z/13Z$.  Usually $g$ and $g^i$ would be conjugate if there would be an element $a\in G$ with $g=a g^i a^{-1}$ and so the conjugacy classes for the rotation would be $\{g^{\pm 1}\}$ and $\{g^{\pm 2}\}$. Is this just one big $K$-conjugacy class because of the second condition
"and $\zeta \mapsto \zeta^r$ is  extensible  to  an  automorphism  of $K(\zeta)$o fixing $K$  elementwise"
which is writen in the mentioned article? Why?
 A: Why is the definition like this?
When K is a field, then K-conjugacy is designed to capture the idea of two group elements being indistinguishable as far as the traces of K-representations.
In other words, $g$ is $K$-conjugate to $h$ in $G$ if whenever $\rho:G \to \operatorname{GL}(n,K)$ is a homomorphism, $\newcommand{\tr}{\operatorname{tr}}\tr(\rho(g))=\tr(\rho(h))$.
It turns out there is a simpler numerical version using the so-called power-map of the character table, and this is the one given by both Pazderski (1985) and Curtis–Reiner's Representation Theory page 306, §42.8.
Since the eigenvalues are $n$th roots of unity for $n=|G|$, the eigenvalues all live in $K[\zeta_n]$. The trace is the sum of the eigenvalues and so $\newcommand{\Gal}{\operatorname{Gal}(K[\zeta_n]/K)}\Gal$ acts on the trace. Of course the matrix $\rho(g)$ has all of its entries in $K$, and since the trace is the sum of the diagonal elements, $\tr(\rho(g))\in K$. In particular, the trace is fixed by every element of $\Gal$.
Elements of $\sigma$ of $\Gal$ are completely defined by what they do to $\zeta_n$, and they must take $\zeta_n \to \zeta_n^r$ for some $r$ with $\gcd(r,n)=1$.
However, a different way of raising the eigenvalues to powers is simply to raise the matrix to a power: if $\rho(g)$ has eigenvalue $\zeta_n^i$, then $\rho(g)^r$ has eigenvalue $\zeta_n^{ri}$.
Hence $$\tr(\rho(g)) = \sigma(\tr(\rho(g))) = \sigma( \zeta_n^i + \zeta_n^j + \dots ) = \zeta_n^{ri} + \zeta_n^{rj} + \dots = \tr(\rho(g^r))$$
Of course, $\tr(ABA^{-1}) = \tr(B)$ so we even have $$\tr(\rho( ag a^{-1} ) ) = \tr(\rho( g^r) )$$ for any $a \in G$ and $r \in \Gal$.
Since this holds for all $K$-representations, we say that $g$ and $a^{-1} g^r a$ are $K$-conjugate.
Specific case
Why are $g$ and $g^{-13}$ (inside the dihedral group of order 10, $g$ order 5) $K$-conjugate where $K=\mathbb{Z}/13\mathbb{Z}$. Well, $\Gal$ is generated by the Frobenius automorphism $\zeta_{10} \mapsto \zeta_{10}^{13}$ and so $g$ and $g^{13}$ are always $K$-conjugate. Since $g$ and $g^{-1}$ are $G$-conjugate, the result follows.
More explicitly $ag^{-1}a^{-1} = g$ is $K$-conjugate to ${(g^{-1})}^r = g^{-13}$.
