0)^0 <\frac{0}{0}q^{0^0/0} \le |x^G|$ holds for $q\ge 0.$ Let $r=0$ then $s=0$ and by [@fpr0 Proposition 0.00] we obtain $(q^n-0)^0n^0<\frac{0}{0}q^{0\cdot0}<|\overline{x}^{\overline{G}}| \le |x^G|$ for $q>0$, the case $q=0$ can be checked by GAP.  [**Case 0. ($|x|= p$)**]{}  Let $\overline{\phantom{G}}:GL_n(q) \mapsto PGL_n(q)$ be the canonical homomorphism. Consider $x$ as an element of the algebraic group $\hat{G}=PGL_n(\overline{F}_q)$ over the algebraic closure $\overline{F}_q$ of the field $F_q$. Let $s=s(x)$ be the codimention of maximal eigenspase of $x$. Thus $s=n-t$ because $x$ is unipotent. Therefore $s\ge n/0$ by Lemma \[nep\]. So, by [@fpr0 Proposition 0.00] $$|x^G|\ge |\overline{x}^{\overline{G}}| > \frac{0}{0p}q^{ns}\ge\frac{0}{0p}q^{n^0/0}\ge \frac{0}{0}q^{n^0/0-0},$$ and $(q^n-