If the automorphism group is cyclic, then the Inner automorphism group is cyclic. But the inner automorphism group is isomorphic to $G/Z(G)$, and if $G/Z(G)$ is cyclic, then it is trivial. Therefore, $G=Z(G)$ so $G$ is abelian. (The argument does not require $G$ to be finite, by the by.)
For the latter:
Prop. If $H\subseteq Z(G)$ and $G/H$ is cyclic, then $G$ is abelian.
Suppose $G/H$ is cyclic, with $H\subseteq Z(G)$. Let $g\in G$ be such that $gH$ generates $G/H$. Then every $x\in G$ can be written as $x=g^kh$ for some integer $k$ and some $h\in H$. Given $x,y\in G$, we have $x=g^kh$ and $y=g^{\ell}h'$, so
$$\begin{align*}
xy &= (g^kh)(g^{\ell}h')\\
&= g^kg^{\ell}hh' &\quad&\text{since }h\in Z(G)\\
&= g^{\ell}g^kh'h\\
&= g^{\ell}h'g^kh &&\text{since }h'\in Z(G)\\
&= (g^{\ell}h')(g^kh)\\
&= yx,
\end{align*}$$
hence $G$ is abelian. QED
For more on what groups can occur as central quotients, see this previous question.
Added. Since you mention you did not know that $\mathrm{Inn}(G)\cong G/Z(G)$, let's do that too:
Define a map $G\to \mathrm{Aut}(G)$ by mapping $g\mapsto \varphi_g$, where $\varphi_g$ is "conjugation by $g$". That is, for all $x\in G$,
$\varphi_g(x) = gxg^{-1}$.
This map is a group homomorphism: if $g,h\in G$, then we want to show that $\varphi_{gh} = \varphi_g\circ\varphi_h$. To that end, let $x\in G$ be any element, and we show that $\varphi_{gh}(x) = \varphi_g(\varphi_h(x))$.
$$\varphi_{gh}(x) = (gh)x(gh)^{-1} = ghxh^{-1}g^{-1}= g(hxh^{-1})g^{-1} = \varphi_g(hxh^{-1}) = \varphi_g(\varphi_h(x)).$$
Therefore, the map $g\mapsto\varphi_g$ is a homomorphism from $G$ onto $\mathrm{Inn}(G)$. By the Isomorphism Theorem, $\mathrm{Inn}(G)$ is isomorphic to $G/N$, where $N$ is the kernel of this homomorphism.
What is $N$? $g\in N$ if and only if $\varphi_g$ is the identity element of $\mathrm{Aut}(G)$, which is the identity; that is, if and only if $\varphi_g(x)=x$ for all $x\in G$. But $\varphi_g(x)=x$ if and only if $gxg^{-1}=x$, if and only if $gx = xg$. So $\varphi_g(x) = x$ if and only if $g$ commutes with $x$. Thus, $\varphi_g(x)=x$ for all $x$ if and only if $g$ commutes with all $x$, if and only if $g\in Z(G)$. Thus, $N=Z(G)$, so $\mathrm{Inn}(G)\cong G/Z(G)$, as claimed.
But what about the converse?
Is false. Take $G=\mathbb{Z}_2\oplus\mathbb{Z}_2$. Or more explicitely $G=\{e, a, b, c\}$ with addition
$$e\mbox{ is neutral}$$
$$a+a=e$$
$$b+b=e$$
$$c+c=e$$
$$a+b=c$$
$$a+c=b$$
$$b+c=a$$
Also known as the Klien four-group. This group is abelian. It is not cyclic because it is of order $4$ while every (nontrivial) element is of order $2$.
Finite (or even finitely generated) abelian groups are all known. If $G$ is a finite abelian group then
$$G\simeq\mathbb{Z}_{p_1^{a_1}}\oplus\cdots\oplus \mathbb{Z}_{p_m^{a_m}}$$
for some (not necessarily distinct) primes $p_1,\ldots, p_m$ and naturals $a_1,\ldots,a_m$.
For the finitely generated case you have to add $\mathbb{Z}^k$ term.
a group was simple and abelian to conclude that it is cyclic of prime order. Why is this true?
Abelian groups have this neat property: every subgroup is normal. Now if $G$ is of order $n$ then by Cauchy's theorem it has a subgroup $H$ of prime order $p | n$. This subgroup is nontrivial. It is proper if $n$ is not prime. If additionally $G$ is abelian then $H$ is normal, hence $G$ is not simple.
So the only possibility for an abelian group to be simple is when it is of prime order. Again Cauchy's theorem implies that such group has to be cyclic and indeed every $\mathbb{Z}_p$ is simple.
Best Answer
Hint:
An automorphism $f$ of $\mathbf Z_8$ maps the generator $\bar 1$ onto another generator, and this image characterises $f$.
Now the generators of $\mathbf Z_8$ are $\;\{\bar 1,\bar 3,\bar 5,\bar 7\}$, hence $\operatorname{Aut}(\mathbf Z_8)$ has order $4$. Check that any automorphism $f$ satisfies $f^2=\text{id}$, and deduce from this relation that $\operatorname{Aut}(\mathbf Z_8)$ is commutative.
Note: if you know that there are, up to an isomorphism, only two groups of order $4$, and that they both are commutative, it's still shorter.