Prime divisor of Automorphism group 2

In the my last question I asked this question: Suppose that G is a non-abelian simple group and $r$ is a prime number such that $r$ not divide $|G|$. Whether $r$ can be prime divisor of |Aut($G$)|?

I received some good examples for my question. Now I want to ask this question:

Is there any method for finding these type of non-abelian simple groups? If there is how?

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One general way to do this as follows. I illustrate with certain groups ${\rm PSL}(n,p^{r}),$ though the method works much more generally. Take the group ${\rm GL}(n,p)$ for some prime $p.$ Take a prime $r$ which does not divide $|{\rm GL}(n,p)|$ and consider the group ${\rm GL}(n,p^{r}).$ This group admits an automorphism $\alpha$ of order $r$ which is induced by the Frobenius automorphism $x \to x^{p}$ of ${\rm GF}(p^{r}).$ Furthermore, the fixed point subgroup od $\alpha$ is ${\rm GL}(n,p),$ which has order prime to $r$ by hypothesis. Since $\alpha$ has order prime $r,$ it follows that $|{\rm GL}(n,p^{r})|$ is also prime to $r,$ since the elements not fixed by $\alpha$ are all in orbits of length $r.$ Now $\alpha$ induces an automorphism of order $r$ of the simple group ${\rm PSL}(n,p^{r}),$ and this simple group certainly also has order prime to $r.$
Thanks, by your answer and the examples I think it is correct for all simple groups of Lie type over a field with $p^r$ elements. Am I right? – user2132 Jan 3 '13 at 11:52
Well, you need to know that $r$ does not divide the order of the group defined over the prime field for this argument to apply. – Geoff Robinson Jan 3 '13 at 20:01