Finding automorphisms and inner automorphisms of $D_3$ I want to find $Aut(D_3)$ and $Inn(D_3)$.  
Suppose $f:D_3 \to D_3$ is an automorphism. Then we must have: 
$\phi(s)$ must have order $2$. So, we must then have $\phi(s) \in \{s,rs,r^2s \}$
$\phi(r)$ must have order $3$. So, we must then have $\phi(r) \in \{r,r^2\}$
Since there are two choices to determines $\phi$, we have $2\times 3 = 6$ choices for $\phi$. My question is, how do I actually describe an automorphism of $D_{3}$? I think I got myself quite confused.
 A: Since $D_3 = \langle r, s \rangle$, then defining $\phi$ on $\{r, s\}$ is all you need: having defined $\phi$ on $r$ and $s$, we can "extend" $\phi$ to all of $D_3$ by defining $\phi(rs) = \phi(r)\phi(s)$, for example. Similarly with the other products of $r$ and $s$.
This means that, of your $6$ potential automorphisms, you just need to check which satisfy $\big(\phi(rs)\big)^2 = \big(\phi(r)\phi(s)\big)^2 = e$. 
If this is the case, you can be sure that $\phi(r),\ \phi(s),$ and $\phi(r)\phi(s)$ satisfy the same relations as $r$ and $s$ so that $\phi(D_3)$ is given by $\langle \phi(r), \phi(s) \mid \phi(r)^3 = \phi(s)^2 = \big(\phi(r)\phi(s)\big)^2 \rangle$ and is thus isomorphic to $D_3$. In other words, $\phi$ is indeed an automorphism.
<EDIT>
To be more concrete, let's say you pick $\phi(s) = rs$ and $\phi(r) = r^2$. Now let's look at how the extension of $\phi$ works. Our goal is to make $\phi$ a homomorphism, which among other things requires $\phi(g^{-1}) = \phi(g)^{-1}$ for all $g \in G$.
So we'll define $\phi(r^{-1}) = \phi(r)^{-1}$. This means we need
$$\phi(r^2) = \phi(r^{-1}) \overset{\text{def}}= \phi(r)^{-1} = (r^2)^{-1} = r.$$
Since $s$ and $\phi(s) = rs$ are both self-inverse, we have nothing to worry about there. So far, we know where $\phi$ sends $r,\ s,$ and $r^2$. We really hope $\phi(1) = 1$, let's see if "extending" gets us that:
$$\phi(1) = \phi(rr^2) \overset{\text{def}}{=} \phi(r) \phi(r^2) = r^2 r = 1 \checkmark$$
Still missing $\phi(rs)$, let's see what extending says:
$$\phi(rs) \overset{\text{def}}{=} \phi(r)\phi(s) = r^2 \cdot rs = s.$$
All that's left is $r^2s$, and you can probably guess what comes next,
$$\phi(r^2s) \overset{\text{def}}{=} \phi(r^2)\phi(s) = r rs = r^2s.$$
So by extending, we know what our function $\phi$ does:

That's the full description of $\phi$. You can play around and make sure that $\phi(xy) = \phi(x)\phi(y)$ to make sure it's truly an automorphism and not a bijection. Our map was designed to be, so it'll all work out, but playing around with a concrete example will help.
</EDIT>

Thinking about inner automorphisms, you have some options. Note that conjugation by any element of $D_3$ not in the center (which is trivial) leads to a nontrivial automorphism. But, perhaps some of these conjugations are really the same!
To deal with this, suppose that conjugation by $x, y \in D_3$ determine the same function $D_3 \to D_3$; that is, $x^{-1}gx = y^{-1}gy$ for all $g \in D_3$. Show that this means $(xy^{-1})^{-1}g(xy^{-1}) = g$, so that in fact $xy^{-1}$ is in the center of $D_3$ which is centerless, so that $x = y$. Conclude that all automorphisms are inner.
For a more highbrow approach, you could appeal to the fact that $G/Z(G) \cong \operatorname{Inn}(G)$ where again centerless $G$ means that $D_3$ must have $6$ inner autormorphisms, but we more or less proved this fancier statement above.
