67th Romanian Mathematical Olympiad

Since $p$ is a prime divisor of the number of automorphisms of $G$, some automorphism $f$ has order $p$. Since $f$ is a permutation of the set $G\setminus \{e\}$, it follows that $f$ is a cycle of length $p$, so $G\setminus \{e\}=\{x,f(x),\dots ,f^{p-1}(x)\}$, whatever $x$ in $G\setminus \{e\}$. On the other hand, $|G|=p+1$ is even, so $G$ contains an element $x_0$ of order $2$. Hence $\text{ord}{f}^k(x_0)=2$, $k=0,\dots ,p-1$, so $x^2=e$ for all $x$ in $G$, and $p+1=2^n$ for some integer $n\ge 2$. Consequently, $p=2^n-1\equiv 3(\mathrm{mod}4)$.