How does the Frobenius act on the prime-to-$p$ $\pi_1(\mathbb{P}^1_{\overline{\mathbb{F_p}}}\setminus \{a_1,…,a_r\})$?

From Grothendieck's work we know that the prime-to-p fundamental group $\pi_1(\mathbb{P}^1_{\overline{\mathbb{F}_p}}\setminus\{a_1,...,a_r\})$ where $a_1,...,a_r \in \mathbb{F}_p$ is isomorphic to the prime-to-p part of the profinite completion of $\langle \alpha_1,...,\alpha_r|\alpha_1...\alpha_r=1\rangle$.

The question is: how does the Frobenius automorphism of $\overline{\mathbb{F}_p}$ act on the prime-to-p $\pi_1(\mathbb{P}^1_{\overline{\mathbb{F}_p}}\setminus\{a_1,...,a_r\})$ where $a_1,...,a_r \in \mathbb{F}_p$?

I don't actually expect an answer. I gather that this is not well understood.

My question is: what is known about it? Where can I read more? And in general any insight about this question is very welcome.

I put a community wiki stamp on this because there's no one right answer.

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Trivially?:) It seems that you will have an action only if your $a_i$ belong to the fixed field of the Frobenius. – Mikhail Bondarko Dec 21 '10 at 0:59
Ah! This is what I meant. Let me fix it. – Makhalan Duff Dec 21 '10 at 1:05
Then why do you think that the action is non-trivial?:) – Mikhail Bondarko Dec 22 '10 at 1:37
Why would it be trivial? I don't think I follow... – Makhalan Duff Dec 22 '10 at 3:29