A problem concerning action via automorphisms This problem asks to show that if $A$ acts on $G$ via automorphisms, where either $A$ or $G$ is soluble and both $A$ and $G$ are finite groups and $G$ is nontrivial, then $G$ possesses an $A$-invariant $p$-subgroup.
My original attempt at a solution was this:
If $G$ is soluble, then $F(G)>1$, in which case any nontrivial Sylow subgroup of $F(G)$, being characteristic all the way up to $G$, will do the trick.
If $A$ is soluble, choose as above some Sylow $p$-subgroup of $F(A)$, say $P$. Certainly $P$ is normal in $A$ and my claim here is that there exists some $P$-invariant Sylow $q$-subgroup of G, regardless of the condition that $p$ divides $|G|$ or not. If $p$ does not divide $|G|$, then a direct appeal to the extended version of the Sylow theorems for coprime actions proves the claim. If $p$ divides $|G|$, then consider the action of $P$ on $Ω = Syl_p(G)$, which is nonempty. Then $|Ω| =$ sum of $P$-orbits, each a $p$-power, while $|Ω| = 1 (mod p)$. Some $P$-orbit must therefore be a singleton set, meaning that $P$ fixes this particular Sylow $p$-subgroup of $G$. Thus, in all cases the claim holds.
Now suppose that the $P$-invariant Sylow subgroup of $G$ is $Q$, where it's possible that $p$ divides $|Q|$. Then for any $u \in P$ and $a \in A$, $aua^{-1} \in P$. Hence $aua^{-1}(Q)=Q$, implying that $a(Q)Qa^{-1}(Q)=Q$. Thus $a(Q)$ is a subgroup of $N_G(Q)$ and also a Sylow $q$-subgroup of $G$, because $a$ (being an automorphism) maps Sylow subgroups to Sylow subgroups. Therefore $a(Q)$ is a member of $Syl_q(N_G(Q))$, the sole member of which is $Q$. Finally $a(Q)=Q$ for all $a \in A$ and we are done.  
The last paragraph contains an error, but I'm not sure what that is. I haven't used anywhere that $P>1$. But if $P$ is trivial, then $Q$ can be an arbitrary Sylow subgroup of $G$, and of course, $Q$ is not necessarily fixed by $A$. 
 A: You have covered the case that $G$ is solvable.
If $A$ is solvable, let $P\lhd A$ be a nontrivial $p$-subgroup of $A$, and consider $C_G(P)$, the fixed points of $P$ acting on $G$.  Since for all $x\in P$ and $a\in A$, we have $x^a\in P$, it's not hard to see $C_G(P)$ is $A$-invariant.  Applying induction on $|G|$ then reduces to the case of $C_G(P)=\lbrace1\rbrace$.
Since $|G|\equiv |C_G(P)|\equiv 1\pmod{p}$ in this case, we have that $|G|$ and $|P|$ are coprime, and thus for every prime $q$ dividing $|G|$, there is a unique Sylow $q$-subgroup $Q$ of $G$ which is $P$-invariant (this is theorem 3.23 in FGT).  Again, because for all $x\in P$ and $a\in A$, we have $x^a\in P$, it follows that $Q$ is actually $A$-invariant.
A: A much stronger claim is false (actually obviously so; I'll not delete in hopes it helps point out why the actual claim should be seen as a generalization of something true):
Let G be the alternating group of degree 5, and let A be the group of inner automorphisms induced by the alternating group of degree 4.  Let p be 3. Then a non-identity p-subgroup P of G has order 3, and is generated by a 3-cycle.  If A normalizes the 3-cycle, then the 3-cycle cannot move the point 5, since P acts regularly, and so A must send the points moved by P to points moved by p.  Hence P is in fact a normal Sylow 3-subgroup of A, which does not exist.
In particular, A is a finite soluble group of automorphisms acting on the finite group G, p is a prime, but G has no non-identity A-invariant p-subgroups.
Coprime action would be enough to fix this, and is Kurzweil–Stellmacher's Proposition 8.2.3 on page 185.
Without coprime action every soluble, non-nilpotent group A = G provides a counterexample, simply because each such group has a non-normal Sylow p-subgroup for at least one prime p.
However, as Derek Holt points out, surely the problem is just suggesting that there is at least one p, not that all p work.  This is very similar to Fitting's theorem in the A = G soluble case: every finite soluble group contains a non-identity normal p-subgroup for at least one prime p.
