I read the article (https://www.researchgate.net/publication/291552840_Invariant_Sylow_subgroups_and_solvability_of_finite_groups )
I don't understand how easy this is:
"The group $G$ is acted on by an automorphism group $A$. If this is the coprime action ($(|A|,|G|)=1$), $G$ always has $A-$invariant Sylow $p-$subgroups for every prime $p$ (dividing the order of $G$) and there exists exactly one $A-$invariant Sylow $p-$subgroup $P$ if and only if $P$ is normalized by the fixed point subgroup, $C=C_G(A)$."
Thank you very much.
This is a very good question indeed and it is non-trivial to prove. Here is the theorem you are after, which is the coprime action variant of the famous Sylow theorems.
Theorem Let a finite group $A$ acting via automorphisms on a finite group $G$ and assume gcd$(|A|,|G|)=1$. For every prime $p$ dividing $|G|$:
$(a)$ There exists an $A$-invariant Sylow $p$-subgroup.
$(b)$ All $A$-invariant Sylow $p$-subgroups are $C_G(A)$-conjugate.
$(c)$ If $P$ is an $A$-invariant subgroup of $G$, then $P$ is contained in some $A$-invariant Sylow $p$-subgroup.
There are several books where this is proved, but a very clear exposition can be found in the book Finite Group Theory of Marty Isaacs. The proof there makes use of a famous and often applied lemma of George Glauberman. Details can be found here (see (3.23)Theorem and (3.24)Lemma).