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Yes, if you mean contradiction to the simplicity of $G$. Let $P$ be a Sylow $p$-subgroup of ${\rm Aut}(G).$ Let $Q$ be a $P$-invariant Sylow $q$-subgroup of $G$ where $q$ is an odd prime divisor of $p-1$ (there is on such as the number of Sylow $q$-subroups of $G$ is certainly not divisible by $p,$ while Then $P$ permutes them,so fixes at least one. The non-identity elements of $Q$ by conjugation (thinking of $P$ as a subgroup of the semidirect product $GP.$ Since there are certainly fewer than $p$ such elements, each orbit must be of length $1$. In other words, $[P,Q] =1$ in the semidirect product. Let $R$ be a $P$-invariant Sylow $2$-subgroup of $G.$ Since $p+1$ is a power of $2$ and $q$ was an arbitrary odd prime divisor of $p-1,$ we have $G = RC_{G}(P).$ Hence $[G,P] = [R,P] \leq R. $ But $[G,P] \lhd G.$ If $P \neq 1,$ then we have $[G,P] \neq 1.$ Since $G$ is simple, $[G,P] = G \leq R.$ But then $G$ is a $2$-group and $|G| = 2,$ which is not the case.