Please verify the following proof or comment on how would you have proven it.
Suppose $q = p^2$ and we have $f: \mathbb{F}_{q} \to \mathbb{F}_{q}$ where $f(a) = a\cdot a^p$
Let $a\cdot a^p = b\cdot b^p$ $\implies$ $a^{p+1} = b^{p+1}$ $\implies$ $a=b$
Hence it is a 1-1 map.
Now $f(a+b) = (a+b)^p\cdot (a+b) = (a^p+b^p)\cdot (a+b) = a\cdot a^p + a\cdot b^p + b\cdot a^p + b\cdot b^p$
The map does not preserve addition. Hence it is not an automorphism.
The mapping $N(a)=a^{p+1}$ is known as the norm map from $\Bbb{F}_{p^2}$ to $\Bbb{F}_p$. It is a homomorphism of multiplicative groups, but, as you observed, does not respect addition.
Its mapping properties can be most easily deduced as follows. The multiplicative group of $\Bbb{F}_q$ (resp. $\Bbb{F}_p$) is cyclic of order $q-1=(p-1)(p+1)$ (resp. cyclic of order $p-1$). Note that for any $a\in\Bbb{F}_{p^2}$ we have $a=a^q=a^{p^2}$. Therefore $$ N(a)^p=(a^{p+1})^p=a^{p(p+1)}=a^{p^2+p}=a^{p^2}a^p=a\cdot a^p=N(a). $$ An element $x\in\Bbb{F}_{p^2}$ belongs to the subfield $\Bbb{F}_p$ if and only if $x^p=x$, so we can conclude that, indeed, $N(a)\in\Bbb{F}_p$ for all $a\in\Bbb{F}_q$.
In a cyclic group $C_{p^2-1}$ raising to power $p+1$ is a homomorphism of groups and a $(p+1)$-to-$1$ mapping. This is basic theory of cyclic groups. If $g$ is a generator of $C_{p^2-1}=\Bbb{F}_q^*$, then an arbitrary element $x=g^i$ for some exponent $i, 0\le i<q-1$. Thus $N(x)=g^{i(p+1)}=1$ if and only if $i(p+1)$ is divisible by $p^2-1$ which is equivalent to $i$ being divisible by $p-1$. There are $p+1$ such choices of $i$, so $$ |\ker N|=p+1. $$ A homomorphism of groups is 1-1 if and only its kernel is trivial, so that is not the case.
For an example consider the case $p=2$. Then $$ \Bbb{F}_4=\{0,1,\alpha,\alpha+1\}, $$ where $\alpha$ satisfies the equation $\alpha^2=\alpha+1$. Here clearly $N(0)=0$ and $N(1)=1$. We can also calculate that $$ N(\alpha)=\alpha^3=\alpha\cdot\alpha^2=\alpha(\alpha+1)=\alpha^2+\alpha=2\alpha+1=1. $$ Also $$ N(\alpha+1)=N(\alpha^2)=N(\alpha)^2=1^2=1. $$ Thus we verified the earlier claim that $\ker N$ has $p+1=3$ elements.
Summary: The mapping respects multiplication but does not respect addition. Its image is exactly the subfield $\Bbb{F}_p$. The element $0$ has a single preimage. All the other elements of $\Bbb{F}_p$ have $p+1$ preimages.