This might be well-known to experts. I was just teaching a course where we went through some parts of Quillen's theorem computing the higher algebraic K-theory of finite fields. Denote by $\mathbb F\_q$ the field with $q$ elements.



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> **Theorem (Quillen)** Let $q=p^k$ for some prime $p$ and $k \in \mathbb N\_{\geq 1}$. $$K\_{2i-1}(\mathbb F\_q) = \mathbb Z/(q^i-1)\mathbb Z \quad \mbox{and} \quad K\_{2i}(\mathbb F\_q)= 0.$$
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Canonically, $K\_1(\mathbb F\_q) = \mathbb F\_q^\times$, so that the action of ${\rm Aut}(\mathbb F\_q)$ on $K\_1(\mathbb F\_q)$ is clear. Since higher Milnor K-theory of finite fields vanishes, this does not help to understand the action on the higher K-groups.



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> **Question:** How does ${\rm Aut}(\mathbb F\_q)$ act on $K\_{2i-1}(\mathbb F\_q)$?
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Quillen's proof relies on the choice of an embedding $\bar {\mathbb F}\_q^\times \to S^1$, which makes the isomorphism somewhat non-canonical, as far as I understand. Anyway, my guess would be that the Frobenius $F(x) = x^p$ acts as $a \mapsto p^i a$ on $\mathbb Z/(q^i-1)\mathbb Z$, but I do not know how to prove this.



Galois action on algebraic K-theory of finite fields
$K(\overline{\mathbb{F}\_p})$ is, after completion at any prime $\ell \neq p$, equivalent to $ku^\wedge\_\ell$. It's true that the equivalence relies on a non-canonical embedding, but the conclusion that the homotopy groups of the $\ell$-completion are a polynomial ring, is independent of the concrete isomorphism. Since the homotopy groups of $K(\overline{\mathbb{F}\_p})^\wedge\_\ell$ are $\ell$-divisible and concentrated in odd degrees, we get
$$
\pi\_{2n} K(\overline{\mathbb{F}\_p})^\wedge\_\ell = \operatorname{Hom}(\mathbb{Q}\_\ell/\mathbb{Z}\_\ell, \pi\_{2n-1}K(\overline{\mathbb{F}\_p})).
$$
We know the Frobenius acts by multiplication by $p$ on $\pi\_1K(\overline{\mathbb{F}\_p})\cong \overline{\mathbb{F}\_p}^\times$, so it acts by multiplication by $p$ on $\pi\_2 K^\wedge\_\ell$, hence by multiplication by $p^n$ on $\pi\_{2n} K^\wedge\_\ell$, hence by multiplication by $p^n$ on $\pi\_{2n-1} K(\overline{\mathbb{F}\_p})$ as you expected (by going over all $\ell\neq p$). The finite field case follows from this by comparison along the canonical map between them.


