Is it the case that for odd primes $p\geq5$, all finite groups with self-normalizing Sylow $p$-subgroups are solvable? The simple group of order 168 shows that this conjecture does not hold for $p=2$. Verret's answer provides a counterexample when $p=3$.
Does this conjecture hold for any odd primes $p\geq5$? If so, is there a proof that does not rely on the CFSG?
Edit: Initially, I thought that I had proved that a minimal counterexample to this conjecture was necessarily simple. Verret's answer shows that this is not the case.
Fix a prime $p$ and let $G$ be a minimal counterexample to the conjecture. Suppose that $G$ is not simple.
1) Let $H$ be a nontrivial normal subgroup of $G$ and let $P$ be a Sylow $p$-subgroup of $G$.
Then $PH$ is a subgroup of $G$ that contains $N_G(P)$ so $N_G(PH)=PH$.
Also, $PH/H$ is a Sylow $p$-subgroup of $G/H$ with
$$N_{G/H}(PH/H)=N_G(PH)/H=PH/H.$$
Then $G/H$ is solvable by the minimality of $G$.
However, $G$ is not solvable so $H$ is not solvable.
2) Let $H$ be a proper normal subgroup of $G$ with $G/H$ simple. Then $G/H$ is cyclic of prime order. If $G/H$ is not cyclic of order $p$ then $H$ contains a Sylow $p$-subgroup of $G$ and thus Sylow $p$-subgroups of $H$ are self-normalizing. The minimality of $G$ provides a contradiction. This shows that $G/H$ is cyclic of order $p$. If $p$ does not divide the order of $H$ then $G\cong H\rtimes_\varphi C_p$ for some homomorphism $C_p\to Aut(H)$. However, Sylow $p$-subgroups of $G$ are self-normalizing so this homomorphism must be fixed-point-free. Then $H$ admits a fixed-point-free automorphism of prime order which contradicts the non-solvability of $H$. In summary, $p$ divides the order of $H$ and $G/H$ is cyclic of order $p$.
Best Answer
$\mathrm{P\Sigma L}(2,27)$ has a self-normalising Sylow $3$-subgroup.
(On the other hand, $\mathrm{PSL}(2,27)$ does not. There is a mistake in the proof that a minimal example must be simple, in part 2): there is no reason for $N_G(P)$ to contain every Sylow $p$-subgroup of $G$.)
According to http://www.ams.org/journals/proc/2004-132-04/S0002-9939-03-07161-2/S0002-9939-03-07161-2.pdf the only counterexamples for odd $p$ occur for $p=3$ and involve $\mathrm{PSL}(2,3^f)$ as a composition factor.
(See also https://arxiv.org/pdf/1503.08907.pdf)