We define the group $D_n$ to be the dihedral group of order $2n$ (equivalent to the group of rotations and reflections on a regular $n$-gon) and $\operatorname{Aut}(G)$ to be the group of automorphisms of a group. Additionally, we say that $\varphi(n)$ is Euler's totient function.
My conjecture is this: $D_n\approx\operatorname{Aut}(D_n)$ if and only if $\varphi(n)=2$. In other words, this is a property only held by the groups $D_3$, $D_4$, and $D_6$. Is this true?
Here is my reasoning. As we know, $D_n=\langle r,s\ |\ r^n=s^2=e, rs=sr^{-1}\rangle$. As for the elements of the automorphism group, we can define two variables, $\rho^i: s\mapsto sr^i$ and $\sigma^j: r\mapsto r^{f(j)}$ for some valid value $j$ and function $f(j)$. Looking at this geometrically, we can think of $\rho^i$ as rotating the line of reflection for $s$ by an incriment of $\frac{i\pi}{n}$, and we can think of $\sigma^j$ as mapping the incriment of rotation from $\frac{2\pi}{n}$ to $\frac{2f(j)\pi}{n}$. However, we need to be careful how we choose $f(j)$. By definition, because $\sigma^j\in\operatorname{Aut}(D_n)$, $\sigma^j$ must be a homomorphism $\sigma^j: D_n\rightarrow D_n$.
As an example of a poorly chosen and invalid $f(j)$, let's say that $f(j)=j$. Now, consider the group $D_6$. Here, we have $\sigma^2: D_6\rightarrow D_6$ such that $r\mapsto r^2$. Let's look at what this does to the set of rotations in $D_6$:
$$1\mapsto1;\ r\mapsto r^2;\ r^2\mapsto r^4;\ r^3\mapsto1;\ r^4\mapsto r^2;\ r^5\mapsto r^4$$
Clearly, this is not an automorphism, as an automorphism is by definition surjective.
In fact, this is a problem we run into whenever we have a mapping $\sigma^j: r\mapsto r^{f(j)}$ and $\gcd(f(j),n)\neq1$. In other words, $\sigma^j$ is only a valid mapping if $f(j)$ and $n$ are coprime. As anyone who has taken an elementary number theory course can tell you, there is a function that tells you how many numbers are coprime to $n$ and what those numbers are, defined exactly as $\varphi(n)$. Therefore, $\sigma^j$ will be an element of $\operatorname{Aut}(D_n)$ of order $\varphi(n)$ defined as $\sigma^j: r\mapsto r^{f(j)}$ where $f(j)$ is the $(j\mod n)$-th element in the set of numbers coprime to $n$ arranged from least to greatest.
All this is to say that $\operatorname{Aut}(D_n)=\langle\rho,\sigma\ |\ \rho^n=\sigma^{\varphi(n)}=e, \rho\sigma=\sigma\rho^{-1}\rangle$. From here, it's easy to see that $D_n\approx\operatorname{Aut}(D_n)$ if and only if $\varphi(n)=2$, meaning $n$ must be either $3$, $4$, or $6$.
Did I cover all my bases here? At the very least I know that $D_5\not\approx\operatorname{Aut}(D_5)$, and the same goes for $D_8$. I feel confident about my proof, save for the fact that $\rho\sigma=\sigma\rho^{-1}$ needs proving. But, sometimes there's some weird one-off case where maybe a certain property is only true for $n=1,2,3,$ and $7349$, so I wanted to double check that I'm not messing anything up. Thank you!
