$p$ is prime and $p^2\large\mid\normalsize|G|$ Hints needed:

Let $p$ be a prime and $G$ a finite group such that $p^2\large\mid\normalsize|G|$ then $p\large\mid\normalsize|\text{Aut}(G)|$.

 A: We know there is a sylow $p$-group with order $p^{n}$, $n\ge 2$. Name this group to be $H$. Consider the inner automorphism of $G$: $$f: H\rightarrow Aut(G): h\rightarrow hgh^{-1}$$
The rest details are "easy" to fill out(see the comments or deleted parts for help). Note that $n\ge 2$ is necessarily since otherwise $\mathbb{Z}_{p}$ has an automorphism group of order $p-1$, so has no automorphisms of order $p$.
A: I don't think you can prove this without distinguishing some cases.
First suppose $G$ is Abelian; then by the structure theorem it contains either a cyclic factor of order $p^k$ with $k\geq2$ or a factor $(\mathbf Z/p\mathbf Z)^2$ (this is where $p^2\mid |G|$ is used), and you can compute the order of their automorphism groups, which inject into $\mathop{\mathrm{Aut}} G$.
If $G$ is not Abelian, and has a non-central element of order a power of $p$, then conjugation is your friend.
You can show that every element of $G$ is a product (within the cyclic group it generates) of an element of order a power of $p$ and an element of order indivisible by $p$. In the case that remains, you may (I think) use this to prove that you've got a direct product decomposition of $G$, and apply the first case.
