Why are automorphisms of $D_{2n}, n \geq 5$ odd, not always inner? The dihedral group of order $2n$ is often presented as $$D_{2n}= \langle r,s: r^n=s^2=1, rs= sr^{-1} \rangle \text{,}$$ where $r$ denote rotations and $s$ denote reflections of a regular $n$-gon. Furthermore, I've read that for $D_6 \cong S_3$, i.e. the symmetric group of permutations on $3$ letters, every automorphism is inner. I would like to know why this is so for this particular case, and why then is not every automorphism inner for $D_{2n}, n \geq 5$, where $n$ is odd? For the first case, $Z(D_6)=1$, i.e. the center of $D_6$ is trivial, so that the group of inner automorphisms $\text{Inn}(D_6) \cong D_6$, but how can I justify that such automorphisms are actually inner?  
 A: For a more combinatorial answer: there is an isomorphism $\dfrac{D_{n}}{Z(D_{n})}\simeq {\rm Inn}\; D_{n}$. Here $D_n$ has generators $r,s$, $r^n=s^2=rsrs=1$, i.e. it has order $2n$. When $n=1,2$ the group is abelian. In this case there are no inner automorphisms.
Let's determine the center of $D_{2n}$. Write $x=sr^k$. Then $$xrx^{-1}r^{-1}=sr^krr^{-k}sr^{-1}=srsr^{-1}=r^{-2}\neq 1$$ unless $n=2$. So no element of the form $sr^k$ is central for $n>2$. For $x=r^k$, $k<n$, we have $r^ksr^{-k}s=r^{2k}=1$ if and only if $n\mid 2k$. If $n$ is odd and larger than $2$, this cannot happen. Hence the dihedral group is centerless in such case. If $n$ is even, say $n=2k$, then the center has order $2$ and equals $\{1,r^k\}$. Can you prove this?
Thus the inner automorphism group, for $n>2$ has order $2n$ (when $n$ is odd) or $n$ (when $n$ is even). But the automorphism group of $D_{2n}$ has size $n\varphi(n)$. This gives you the desired claim.
A: With the risk of repeating what was said before...
$D_{2n}$ is the group of symmetries of a regular polygon with $n$ sides (and is sometimes called $D_n$... but for us $D_{2n}$ has $2n$ elements).
In the presentation of $D_{2n}$ with $r$ and $s$, denote $t:=s r$. Then the equality $rs = sr^{-1}$ is equivalent to (using also $s^2 = 1$) $sr sr = 1$, that is, $t^2=1$. Moreover, $st=r$ satisfies $(st)^n=1$. Hence, an equivalent presentation of $D_{2n}$ is $\langle s,t \ \colon \ s^2 = t^2=1, (st)^n=1\rangle$. 
To obtain the presentation of $D_{2n}$  from above we can take $s$, $t$ to be reflections of the regular polygon with respect to two axes that form an angle of $\frac{\pi}{n}$. Then $st$ is a rotation of angle $\frac{2\pi}{n}$. The order of $s$, $t$ and $st$ are respectively, exactly $2,2,n$.
A bit more about the presentation: if we have a group $D$ generated by two elements $s$, $t$ of order $2$ and $N$ is the order of $st$ then $D$ is naturally isomorphic to $D_{2N}$. 
Assume  now on that $n$ is odd. There are $n$ elements of order $2$ in $D_{2n}$, the $n$ reflections with respect to axes passing through a vertex of the polygon and its center. Any two such reflections are conjugate elements, since any such axis can be taken to another axis by a transformation of the polygon. 
Here is how to produce all the automorphisms of $D_{2n}$. We need to find the images $s_1$ and $t_1$ of $s$ and $t$. These have to be elements of order $2$ and the order of $s_1 t_1$ is $n$. 
Assume  now on that $n$ is odd. There are $n$ elements of order $2$ in $D_{2n}$, the $n$ reflections with respect to axes passing through a vertex of the polygon and its center. Any two such reflections are conjugate elements, since any such axis can be taken to another axis by a transformation of the polygon. 
Then $s_1$, $t_1$ are also reflections with respect to two axes of symmetry of the polygon.  Since the order of $s_1t_1$ is $n$, the angle between these axes is $\frac{k \pi}{n}$ where $\gcd(k,n)=1$.  
The automorphism $s\mapsto s_1$, $t\mapsto t_1$ is a conjugation by $\phi$ if and only if the tranformation $\phi$ of the polygon takes the axis of $s$ to the axis of $s_1$ and the axis of $t$ to the axis of $t_1$. But  such a $\phi$ exists if and only if the angles between these pairs of axes are the same. This may not be always be true  for odd $n>3$. ( and also for even $n$ > 4..). 
Here is a picture for the group of a regular pentagon.

The red reflections map to blue ones. This gives an automorphism of $D_{10}$ which is not inner. 
A: It is easy enough to write down all of the automorphisms of $D_{2n}$. The entire automorphism is determined by what it does to the generators. $r$ has order $n$ and must be sent to another element that also has order $n$. Setting $D_2$ and $D_4$ aside as special cases, this means that $f(r)$ must be $r^k$ for some $k$ that is coprime to $n$, and $f(s)$ must then be chosen in $D_{2n}\setminus\langle r\rangle$ -- that is the elements of the form $sr^i$.
With a bit of calculation one sees that each such choice of $f(r)$ and $f(s)$ indeed gives rise to an automorphism.
However, in order for $f$ to be an inner automorphism, it must send $r$ to some conjugate of $r$. But the conjugacy class of $r$ is $\{r,r^{-1}\}$ --  which is easy to see, because this set is closed under conjugating it by $r$ (does nothing) as well as $s$ (swaps $r$ and $r^{-1}$).
So the automorphisms that take $r$ to $r^k$ can only be inner when $k=1$ or $k=n-1$. This means that there must be non-inner automorphisms as soon as $n$ is large enough that there are numbers that are coprime to $n$ but neither $1$ nor $-1$ modulo $n$ -- in other words there are non-inner automorphisms whenever $\varphi(n)>2$ (where $\varphi$ is the totient function).
And the only positive integers with $\varphi(n)\le 2$ are $\{1,2,3,4,6\}$. For example, if $n$ is odd and $\ge 5$, there's at least four different totatives, namely $1$, $n-1$ and $\frac12(n\pm1)$.

It turns out that $D_{2n}$ also has non-inner automorphisms for $n=2,4,6$. Namely, there's an automorphism that sends $r$ to $r$ and $s$ to $sr$, but when $n$ is even, the conjugates of $s$ are all of the form $sr^{2j}$, so again this automorphism cannot be inner.
To complete the list, $D_2$ has only the trivial automorphism, which is inner.
