In this comment from back in 2013, it is claimed that
$${\rm Aut}(D_{2^n})\cong{\rm Aut}(Q_{2^n})$$
for $n\ge 4$, where
$$D_{2^n}\cong \langle r,s\mid r^{2^{n-1}}, s^2, srs=r^{-1}\rangle$$
is the dihedral group of order $2^n$ and
$$Q_{2^n}\cong\langle x,y\mid x^{2^{n-1}}, y^2=x^{2^{n-2}}, y^{-1}xy=x^{-1}\rangle$$
is the generalised quaternion group of order $2^n$ (defined for $n\ge 3$). (For a question of mine on generalised quaternion groups, see here.)
I would like to prove that claim. Please would you help me?
I know that automorphisms are determined by how they behave on generators. Since the presentations above are very similar, I'm not surprised by the theorem.
Perhaps we could use the $N/C$ theorem. Here is the statement:
Theorem: Let $H\le G$ as groups. Then $N_G(H)/C_G(H)$ is isomorphic to a subgroup of ${\rm Aut}(H).$
For a proof, see Gallian's "Contemporary Abstract Algebra (Eighth Edition)", Example 15, page 217.
Here
$$N_G(H)=\{ x\in G\mid xHx^{-1}=H\}$$
and
$$C_G(H)=\{ x\in G\mid xhx^{-1}=h\text{ for all }h\in H\}.$$
So if we let, say, $G=S_{2^n}$, we have some $K,L\le G$ such that $K\cong D_{2^n}$ and $L\cong Q_{2^n}$. The $N/C$ theorem requires that we find $N_G(K), N_G(L), C_G(K), C_G(L)$. If they're all suitably compatible (for lack of a better phrase), then it might follow that
$${\rm Aut}(K)\cong {\rm Aut}(L).$$
I have one condition I would like to add: please do not use the holomorph
$${\rm Hol}(\Bbb Z_{2^{n-1}})$$
because I aim to use the result in question to better understand the fact that
$${\rm Aut}(Q_{2^n})\cong {\rm Hol}(\Bbb Z_{2^{n-1}})$$
for $n>3$. (Proving that isomorphism is Exercise 5.3.4 of Robinson's "A Course in the Theory of Groups (Second Edition)".)
Best Answer
Here is one general way of computing automorphism groups.
Consider for example $G = \langle r,s \mid r^{2^{n-1}} = s^2 = 1, srs^{-1} = r^{-1} \rangle$.
Any automorphism $\phi: G \rightarrow G$ is determined by $\phi(r)$ and $\phi(s)$. Furthermore $\phi(r)$ and $\phi(s)$ must satisfy the same relations, i.e. $$\phi(r)^{2^{n-1}} = \phi(s)^2 = 1,\ \phi(s)\phi(r)\phi(s)^{-1} = \phi(r)^{-1}.$$
Conversely, suppose that you can find generators $x,y \in G$ of $G$ such that they satisfy the same relations: $$x^{2^{n-1}} = y^2 = 1,\ yxy^{-1} = x^{-1}.$$
Then by the universal property of free groups, there exists a surjective homomorphism $\phi_{x,y}: G \rightarrow G$ with $r \mapsto x$ and $s \mapsto y$. Since $G$ is finite, the map $\phi_{x,y}$ must also be an automorphism.
So to determine $\operatorname{Aut}(G)$, it will suffice to find pairs $(x,y)$ which generate $G$ and have the same relations as $(r,s)$. Check that they are $x = r^d$ with $d$ odd and $y = r^e s$ with $e$ integer. This tells you precisely what the automorphisms of $G$ are.
Then you want to do the same thing with $H = \langle r,s\mid r^{2^{n-1}} = 1, s^2=r^{2^{n-2}}, s^{-1}rs=r^{-1} \rangle$. Assuming $n > 3$, the automorphisms of $H$ correspond to the same pairs $(x,y) = (r^d, r^e s)$ as for $G$. From this it should be clear that $\operatorname{Aut}(G) \cong \operatorname{Aut}(H)$.