Let $G$ be an unsolvable group of order $(p+1)t$, where $t\mid (p1)/2$. Is it true that $p$ does not divide $Aut(G)$?

I believe the answer is yes. Let $\alpha$ be an automorphism of $G$ of order $p$. By the Frattini argument, for any prime $q$ dividing $G$, a Sylow $q$subgroup of $G$ is normalized by an element of order $p$ in the semidirect $S$ product of $G$ with $\langle \alpha \rangle$. Since all elements of order $p$ are conjugate in $S$, it follows that $\alpha$ normalizes some Sylow $q$subgroup $Q$. But, since the only prime that could divide both $p+1$ and $t$ is 2, we have $p > Q$ whenever $q>2$ and hence $\alpha$ centralizes $Q$. We also have $p > Q$ for $q=2$, and hence $\alpha=1$, except in the case when $p$ is a Mersenne prime. So we have reduced to the case when $p$ is a Mersenne prime, and hence $t$ is odd. Now $\alpha$ centralizes a subgroup $T$ of $G$ of order $t$ and acts nontrivially on a Sylow 2subgroup $Q$, which has order $p+1$ (and must be elementary abelian). In the semidirect product $S$, we have $[\langle \alpha \rangle, G] = [\langle \alpha \rangle, TQ] = [\langle \alpha \rangle, Q] = Q$, so $Q$ is normal in $G$. Hence, since the odd order subgroup $T$ is solvable, $G$ is solvable, contradiction. 


It is not an answer yet, just the start. Perhaps somebody else can finish. Suppose that $p$ divides the order of $Aut(G)$. Then $G$ admits an automorphism $\sigma$ of order $p$. Let $S$ be the subgroup of fixed points of $\sigma$. Then $S\equiv (p+1)t \equiv t \mod p$. Hence $S=t+rp$ for some $r$. We have that $S$ divides $(p+1)t$. Suppose that $r>0$ (it cannot be smaller than $0$). Let $k$ be the GCD of $t$ and $r$, $r=kr', t=kt'$. Then $t'+r'p$ divides $(p+1)t'$. Since $t'+r'p$ is coprime with $t$, we get that $t'+r'p$ must divide $p+1$. Hence $t'=1, r'=1$, $t=r$. Thus $S=t+tp=G$, a contradiction. Hence $r=0$ and $S=t$. Thus the fixed point subgroup of $\sigma$ must have order $t$ and the number of nontrivial orbits of $\sigma$ is $t$ as well. 

