Show that $\operatorname{Aut}(C_{14})$ is a cyclic group. In the context of abstract algebra and groups automorphism, I need help to understand the solution to this problem:

Show that $\operatorname{Aut}(C_{14})$ is a cyclic group. 

Here's the solution (I'm directly translating from Greek handwritten notes so I hope there are no mistakes):
Since $C_{14}$ is cyclic with $14$ elements, $C_{14} \cong Z_{14}$. Hence, it suffice to show that $\operatorname{Aut}(\mathbb{Z}_{14})$ is cyclic. Also, $\mathbb{Z}_{14} = \{0, 1, 2, 3, \ldots 13\} = \langle 1 \rangle = \langle 3 \rangle = \langle 5 \rangle = \langle 9 \rangle = \langle 11 \rangle = \langle 13 \rangle$. Hence, there will be $6$ automorphism of $\mathbb{Z}_{14}$, specifically 
$$f_1(1) = 1, \,\, f_2(1) = 3, \,\, f_3(1) = 5, \,\, f_4(1) = 9, \,\, f_5(1) = 11, \,\, f_6(1) = 13.$$ 
Therefore, $\left|\operatorname{Aut(\mathbb{Z}_{14})}\right| = 6$.

First the first part of the solution, if I understand the theory correctly, we are using two main "ingredients". The first one is that if a group $G$ is cyclic with $n$ elements, then $G$ is isomorphic to $(\mathbb{Z}_n,+)$. The second proposition we use is that for a group $G$ generated by $a$, i.e. $G =  \langle a \rangle$, if $b$ is a generator of $G$, there is a unique automorphism $f:G \to G$ such that $f(a) = b$. Is my understanding correct so far? 

Now for the second part of the solution which I really don't understand:
Immediately, $\operatorname{ord} f_1 = 1 |$ identity element of $(\operatorname{Aut(\mathbb{Z}_{14})}, \circ)$. 
We have that $\operatorname{ord} f_2 | 6 \implies \operatorname{ord} f_2 = 2 \,\, \text{or} \,\, 3 \,\, \text{or} \,\, 6$. 
Then, $f_2^2(1) = f_2(f_2(1)) = f_2(3) = f_2(1+1+1) = 3f_2(1) = 3.3 = 9$.
Then, $f_2^3(1) = f_2(f_2(3)) = 9 = 9f_2(1) = 9.3 = 27 = 13.$
Therefore, $\operatorname{ord} f_2 = 6 \left|\operatorname{Aut(\mathbb{Z}_{14})}\right|$ and $\operatorname{Aut(\mathbb{Z}_{14})}$ is cyclic, implying $\operatorname{Aut(\mathbb{C}_{14})}$ is cyclic.
I cannot understand anything from the second part of the solution. I think it is very misleading. He is using some divisibility argument and composition of functions but I don't understand what he is actually calculating, why it is necessary and how he is able to conclude from those calculations. I'm looking for an answer that will help me understand in detail the second part of the solution, possibly with the background theory that is needed to solve this kind of problem.
 A: Your understanding of the first part looks correct.
For the second part, first recall that $\operatorname{Aut}(\mathbb Z_{14})$ is the group of automorphisms of $\mathbb Z_{14}$, where the group operation is composition of functions, so it should not be surprising to see that functions are being composed in the author's solution. The group identity is the identity map, which we have labeled $f_1$ in the first part of this problem.
The author uses Lagrange's theorem to observe that the order of $f_2$ in $\operatorname{Aut}(\mathbb Z_{14})$ must divide $|\operatorname{Aut}(\mathbb Z_{14})| = 6$, hence the order of $f_2$ must be one of $1,2,3,6$.
The order of $f_2$ certainly isn't $1$ because $f_2$ is not the identity (it does not map $1$ to itself).
Then the author explicitly computes $f_2^2(1)$ and $f_2^3(1)$, and since neither of these is $1$, it follows that neither $f_2^2$ nor $f_2^3$ is the identity, hence the order of $f_2$ is neither $2$ nor $3$.
(In fact, since $f_2^2(1) = 9$ and $f_4(1) = 9$, it follows that $f_2^2 = f_4$. Similarly, $f_2^3(1) = 13 = f_6(1)$, so $f_2^3 = f_6$.)
Therefore, the only remaining possibility is that the order of $f_2$ must be $6$, so $f_2$ generates $\operatorname{Aut}(\mathbb Z_{14})$, proving that the latter is cyclic.
