Outer automorphism of a finite simple group which is isomorphic to a subgroup of $S_p$ Here is a statement having a proof that involved the CFSG.
Let $p$ be a prime, and $S$ be a nonabelian finite simple group such that $S$ is isomorphic to a subgroup of $S_p$ with $p\mid |S|$. Then $\mathrm{Out}(S)$ is a $p'$-group (i.e. $p\nmid |\mathrm{Out}(S)|$).
Question Is there a CFSG-free proof for this statement?
 A: I think that this may be a result of H. Wielandt, and that, if so, it was probably done without CFSG. I do not remember Wielandt's proof (if my memory of the existence of such a proof is correct), but I think it can be done without CFSG if $p$ is not a Fermat prime, by the following argument, which requires some block theory, and a Theorem of E. Shult.
$\newcommand\card[1]{\lvert#1\rvert}\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Out{Out}$Suppose that $p^{2}\mid \card{\Aut(S)}$ and that $p\mid \card S$. Then there is a subgroup $H$ of $\Aut(S)$ with $[H:S] = p$. Notice that $H \neq O^{p}(H)= S$. Let $P$ be a Sylow $p$-subgroup of $H$, so that $\card = p^{2}$. Then $P \not \leq H^{\prime}$, and $P$ is Abelian. Let $N = N_{H}(P)$ and $T$ be a complement to $P$ in $N$. Then by standard results on coprime actions, we have $P = [P,T] \times C_{P}(T)$, and we have $P \neq P \cap H^{\prime} = [P,T]$. Thus $P$ is not cyclic, and must be elementary Abelian of order $p^{2}$.
Let $Q = C_{P}(T)$, which we now know to have order $p$. Notice that the principal $p$-block of $S$ is $Q$-stable. The principal $p$-block of $S$ contains $e + \frac{p-1}{e}$ irreducible characters, by theorems of R. Brauer, where $e$ is a divisor of $p-1$ (and also $e >1$). Note that this number is at most $p$. Let $R = P \cap S$, which also has order $p$.
Notice that $C_{S}(R) = R$, since $S$ is isomorphic to a subgroup of the symmetric group $S_{p}$. Thus the principal $p$-block of $S$ is the only $p$-block of $S$ of positive defect, and contains all irreducible characters of $S$ of degree prime to $p$ (and, by Brauer's results, all irreducible characters in the principal  $p$-block of $S$ do have degree prime to $p$).
Consideration of the permutation character of degree $p$ of $S_{p}$ shows that $S$ has a non-trivial complex irreducible character of degree less than $p$, say $\chi$,
and we note that $\chi$ is  then faithful (of degree greater than $1$), since $S$ is simple. Then $\chi$ is in the principal $p$-block of $S$, since it has degree prime to $p$.
Since the principal $p$-block of $S$ contains at most $p$ irreducible characters, and the principal $p$-block of $S$ is $Q$-stable (containing the trivial character), we deduce that each irreducible character in the principal $p$-block of $S$ is $Q$-stable, and hence each extends (in exactly $p$ ways) to an irreducible character of $H$. It follows that $\chi$ extends to a faithful irreducible character $\chi^{\ast}$ of $H$.
Let $x$ be a generator of $Q$ other $H$-conjugates (for two elements of $P$ which are conjugate in $S$ are already conjugate in $N_{H}(P)$, as $P$ is Abelian (by an argument of Burnside)).
Hence, in the terminology of E. Shult, $x$ satisfies the ucc. Furthermore, since $\chi^{\ast}$ is a faithful irreducible character of $H$ of degree less than $p$, it follows by a theorem of Hall–Higman–Shult  that $x$ centralizes every $p^{\prime}$-group which it normalizes (this uses the fact that $p$ is not a Fermat prime).
By a theorem of E. Shult (circa 1969— I will try to locate a reference), an element of order $p$ in a finite which satisfies the ucc, and has the additional property that it centralizes every $p^{\prime}$-group which it normalizes, is central in that group.
The result of Shult does not require CFSG, but eventually reduces to the famous theorem of Frobenius about groups which have trivial intersection with all their other conjugates in a finite group.
Note that if $G$ is a minimal counterexample to this theorem of Shult, and $y$ is a non-central element of order $p$ which satisfies the ucc, and centralizes every $p^{\prime}$-group which it normalizes, then $y$ is central in any maximal subgroup $M$ of $G$ with $y \in M$. Hence $y$ is contained in a unique maximal subgroup $M$ of $G$, and we have $M = C_{G}(y)$. It is easy to check that $G$ has no non-trivial normal subgroup of order prime to $p$ and that (since $y$ satisfies ucc), any two elements of $M$ which are conjugate in $G$ are already conjugate in $M$. Furthermore, $C_{G}(m) \leq M$ whenever $1 \neq m \in M$, so we find that $M \cap M^{g} = 1$ for all $g \in G \backslash M$. Hence the Theorem of Frobenius applies,
yielding a normal subgroup of $G$ with $G = MK$ and $K \cap M = 1$. Then $K$ has order prime to $p$, and (by hypothesis) is centralized by $y$, a contradiction (I have essentially reproduced the proof of Shult here).
This brings out the fact that the question is related to odd analogues of Glauberman's $Z^{\ast}$-theorem. Such analogues are known to hold, but for odd $p$, they currently require CFSG. It follows  from the “$Z_{p}^{\ast}$-theorem” that if $X$ is any finite simple group with a Sylow $p$-subgroup of order $p$, then $\Out(X)$ has order prime to $p$. However, the hypotheses of the question here are rather stronger than that.
A: Let $p$ be a prime, and let $G$ be a non-abelian simple group with $p \mid |G|$. Suppose that $G \leq S_p$ and that $G$ is transitive.
By a theorem of Burnside a non-solvable transitive group of prime degree is $2$-transitive, so $G$ is $2$-transitive.
Let $B \leq G$ be such that $|B| = p$. Then $G = AB$, where $A$ is a point stabilizer.  Note that $A \cap B = 1$, so $A$ is a Hall subgroup.
Now it seems to be a theorem of Wielandt that in this case $Aut(G)/G$ is cyclic of order dividing $p-1$. This is according to the following paper of Cameron (I haven't been able to find the paper by Wielandt), see §3 there.

Cameron, Peter J. On groups with several doubly-transitive permutation representations. Math. Z. 128 (1972), 1–14.

The paper by Wielandt is

Wielandt, Helmut. On automorphisms of doubly transitive permutation groups. 1967 Proc. Internat Conf. Theory of Groups (Canberra, 1965) pp. 389–393 Gordon and Breach, New York

You could also use the proofs given by Cameron, which do not need CFSG.
Denote the image of $N_{S_p}(G)$ in $Aut(G)$ by $P(G)$, this is called the group of permutation automorphisms. Then Theorem 3 in Cameron shows that $P(G)$ is normal in $Aut(G)$ and $Aut(G)/P(G)$ is an elementary abelian $2$-group.
Next we can show that $P(G)/Inn(G)$ is cyclic of order dividing $p-1$, which gives you the desired result. For this let $R = N_{S_p}(G)$. By the Frattini argument $R = GN_R(B)$. Then $$\frac{P(G)}{Inn(G)} \cong \frac{R}{GC_{S_p}(G)} = \frac{GN_R(B)}{GC_{S_p}(G)} \cong \frac{N_R(B)}{GC_{S_p(G)} \cap N_R(B)}.$$
Because $B \leq GC_{S_p(G)} \cap N_R(B)$, it follows that $P(G)/Inn(G)$ is isomorphic to a quotient of $N_R(B)/B$. Now $B$ is self-centralizing in $S_p$, so $N_R(B)/B$ is isomorphic to a subgroup of $Aut(B)$, hence cyclic of order dividing $p-1$.
