Computing the order of n-folded automorphisms

1. Jul 20, 2015

jackmell

I'd like to find $\operatorname{aut}^2 \mathbb{Z}_n^*$ and would first like to just compute it's order. However, I'm pretty sure in general $\operatorname{aut}\mathbb{Z}_n^*$ is non-abelian so can't apply or extend the formula for computing $\big|\operatorname{aut}\mathbb{Z}_n^*\big|$ to the twice-folded case.

Is there a formula for computing $\big|\operatorname{aut}^2 \mathbb{Z}_n^*\big|$?

For example, empirically I find $\big|\operatorname{aut}^2\mathbb{Z}_{15}^*\big|=8$. I was wondering if someone could help me prove this algebraically and then perhaps help me extend it to the first 500 groups? I should point out if $\operatorname{aut}\mathbb{Z}_{15}^*$ was abelian, then it would be isomorphic to $\mathbb{Z}_{2}\times \mathbb{Z}_{2^2}$ and then show it's order is 8 but I don't think $\operatorname{aut}\mathbb{Z}_{15}^*$ is abelian.