The automorphism group of $C_{17}$ is isomorphic to $C_{16}$. If you want to embed $C_2\cong\langle y \rangle$ into it, then letting $C_{16}\cong \langle x \rangle$, you can just send $y\mapsto x^8$. Of course, this isn't very satisfying, and you probably want to actually construct the automorphism itself; this requires some explanation.
If you've taken number theory, you've seen $(\mathbb{Z}/17\mathbb{Z})^\times$, the group of units modulo $17$. If you've run into finite fields, you know that the additive group of $\mathbb{F}_{17}$ is isomorphic to $\mathbb{Z}/17\mathbb{Z}$ and the multiplicative group of $\mathbb{F}_{17}$ is $(\mathbb{Z}/17\mathbb{Z})^\times$. If you haven't seen either of those yet, it turns out that if you multiply the integers in a cyclic group of prime order rather than adding them, you get a group as long as you remove $0$ (since it has no multiplicative inverse). You can read about this in any elementary text on modular arithmetic (including Wikipedia).
What I am getting at is that there is a rather elegant description of the automorphism group of a cyclic group by multiplications of its elements. If we take an $a\not= 0\in \mathbb{Z}/17\mathbb{Z}$, we can define the function $\theta_a:\mathbb{Z}/17\mathbb{Z}\rightarrow \mathbb{Z}/17\mathbb{Z}$ by $\theta_a(x)=ax$. Noting that $\theta_a(x+y)=a(x+y)=ax+ay=\theta_a(x)+\theta_a(y)$ and that $\theta_a(x)=ax=0$ implies that $x=0$, we see that this is an automorphism of $\mathbb{Z}/17\mathbb{Z}$.
We can show that such automorphisms form a group under composition. $(\theta_a\theta_b)(x)=\theta_a(\theta_b(x))=a(bx)=(ab)x=(\theta_{ab})(x)$ so $\theta_a\theta_b=\theta_{ab}$. (Here, you can see the isomorphism to $(\mathbb{Z}/17\mathbb{Z})^\times$ popping up in the subscript: the map $\theta_a\mapsto a$.) The rest of the group axioms are verified easily. It turns out that this group of $\theta_a$'s constitutes the entire automorphism group of $\mathbb{Z}/17\mathbb{Z}$.
So, finding a nontrivial embedding of $C_2$ into $C_{17}$ reduces to finding an element $a\in (\mathbb{Z}/17\mathbb{Z})^\times$ with order $2$. Noting that $16\equiv-1\mod{17}$, you see that the map is $\phi:C_2\rightarrow \text{Aut}{C_2}$ by $\phi:y\rightarrow \theta_{16}$. Of course, this method will work to find the automorphism group of $C_p$ for any prime $p$, and the automorphism group of any cyclic group $C_n$ can be described similarly as long as the $a$'s in $\theta_a$ correspond to units modulo $n$. (For a good time, try forming the semidirect product of $C_n$ by $\text{Aut}{C_n}$ - you now know how to make a holomorph.)