Let $(G, \cdot)$ be a finite group, $|G| = n \in \mathbb{N}^*$such that $\forall p \mid n, p$ prime number, there exists $ a \in G$ with order equal with $p$ and $f(a) = a, \forall f : G \to G $ automorphism. Show that $n$ is not square-free.
I have assumed that $n$ is square-free and then tried to obtain a contradiction buy using the automorphisms $f_i(x) = ixi^{-1}, i \in G$, but I didn't manage to solve the problem.
Let $p\mid n$. By assumption, we find $a=a(p)$ that is invariant in particular under inner morphisms, i.e., $ax=xa$ for all $x\in G$. It follows that the subgroup $A$ generated by all $a(p)$ is abelian. Note that $A$ has the automorphism $\iota\colon x\mapsto x^{-1}$. We conclude that $G\ne A$ or that $\iota=\operatorname{id}_A$. In the first case, $n$ is not square-free because $|A|$ is the product of all prime divisors of $n$, hence $n$ contains one of these primes multiply.
In the second case, $|A|=2$, hence $n$ is a power of $2$. This makes $n$ divisible by the square $4$, or we have $n=2$. Indeed, $n=2$ is the only exception to the claim: The non-trivial element of the group of order $2$ does fulfil the conditions for the element $a$ for $p=2$.