Stuck on constructing my semi direct product I want to work out $H \rtimes Q$, where $H = C_{17}$ and $Q = C_{2}$. What this means is that I want to work out the groups that map $\theta: C_2 \rightarrow Aut(C_{17})$.
$Aut(C_{17}) \cong C_{16}$. I know that 2 divides 16 and from here I get two SDP's. One is the direct product, $C_{17} \times C_2$ and the other is the SDP I have to find. What I have said is this:
If we denote $H = \langle a | a^{17} = 1 \rangle, Q = \langle b | b^2 = 1 \rangle$ and $Aut(C_{17}) = \langle \mu | \mu^{16} = 1 \rangle$, then I want to find some $\mu \in Aut(C_{17})$ which will map elements of order $2$ in $Aut(H)$ to elements of order $2$ in $H$, i.e some 
$$\mu \in Aut(C_{17}) \mathrm{\, such \, that\, } \mu : a \mapsto a^k \mathrm{\,goes \, to\,} \mu^{2} : a \mapsto a^{k^2} $$
In other words, I want the $k$'s such that $k^2 \equiv 1 \mod 17$. Is this right so far? I'm stuck on how I go about trying to answer the question now.
 A: So I had assumed previously that you knew what an automorphism group is.  Perhaps you need some elaboration on this.  I'm going to go through this particular example step by step as thoroughly as I can.
$\text{Aut}(C_{17})$ is a cyclic group of order $16$, but it is not $\mathbb{Z}_{16}$ (just isomorphic to it).  It is generated by the bijective homomorphism $\alpha:C_{17}\rightarrow C_{17}$ defined by $\alpha(x)=x^3$.  So $\alpha^2$ is the map $x\mapsto x^{3^2}$, $\alpha^3$ is the map $x\mapsto x^{3^3}$, etc.  Because of number theory stuff, $\alpha^{16}$ is the map $x\mapsto x$, which is the identity homomorphism $\text{id}$ on $C_{17}$, and no other $\alpha^{i}$ is the identity for $i\leq 15$.  So $$\text{Aut}(C_{17})=\{\text{id},\alpha,\alpha^2,\ldots, \alpha^{15}\}.$$
We want to find a homomorphism $\theta:C_2\rightarrow \text{Aut}(C_{17})$ to define the semidirect product.  You already found that the homomorphism $\theta_1:C_2\rightarrow \text{Aut}(C_{17})$ defined by $\theta_1(x)=\text{id}$ yields $C_{17}\rtimes_{\theta_1}C_2=C_{17}\times C_2$. This is one of the two possible semidirect products. We find the other one by noticing that the order of $a$ (the generator of $C_2$) is $2$, so it needs to map to the automorphism $\alpha^8$, because $(\alpha^8)^2=\alpha^{16}=\text{id}$. So let $\theta:C_2\rightarrow \text{Aut}(C_{17})$ be defined by $\theta(a)=\alpha^8$.
Remember that $x^3$ means $3x$ in $\mathbb{Z}_{17}$, since the operation is addition.  The relation to DonAntonio's comment is that $x^{3^8}$ means $3^8x$ in $\mathbb{Z}_{17}$.  But $3^8\equiv 16 \pmod{17}$, i.e. $3^8\equiv -1 \pmod{17}$.  So the map $\alpha^8$ is the map $x\mapsto -x$.
You asked in the comments if you should always use $x\mapsto -x$ for an automorphism of order $2$.  The answer is no, because it is not true in general that this map is a homomorphism.  In fact, it is if and only if the group is abelian.  You should prove this real quick.  (Hint: Let $\phi:x\mapsto x^{-1}$. $b^{-1}a^{-1}=\phi(ab)=\phi(a)\phi(b)=a^{-1}b^{-1}$.)  So only with abelian groups can you use this as an automorphism of order $2$ (though it may not be the only automorphism of order $2$).
So what does this mean for your semidirect product $C_{17}\rtimes_{\theta} C_2$?
We proved yesterday that $ba=a\theta(a^{-1})(b)$.  In $C_2$ $a=a^{-1}$, so $ba=a\theta(a)(b)$.  As we determined above, $\theta$ maps $a$ to the automorphism $x\mapsto x^{-1}$, so $\theta(a)(b)=b^{-1}$, so $ba=ab^{-1}$.  We can rewrite that as $b^a=b^{-1}$, yielding the familiar group presentation $$C_{17}\rtimes_{\theta} C_2=\langle a,b|a^2,b^{17},b^a=b^{-1}\rangle$$ which we recognize as the dihedral group $D_{17}$.
