Is $\operatorname{Aut}(D_{12})\simeq D_{12}$? Let $D_{12}$ be the dihedral group of order 12. Then
$$|\operatorname{Aut}(D_{12})|=6\phi(6)=12=|D_{12}|,$$
and the standard method of proof for
$$\operatorname{Aut}(D_6)\simeq D_{6}\qquad\mbox{and}\qquad \operatorname{Aut}(D_8)\simeq D_{8}$$
seems to also work for
$$\operatorname{Aut}(D_{12})\simeq D_{12}.$$
But in this article (p.461, 14th line from the top), it says that $n=3$ and $n=4$ are the only numbers for which
$$\operatorname{Aut}(D_{2n})\simeq D_{2n}.$$
Could this be an error? Or am I missing something?

The article is wrong. First, the automorphism group is definitely $D_{12}$: here is Magma code that proves it.
> G:=DihedralGroup(6);
> A:=AutomorphismGroup(G);
> A:=PermutationGroup(A);
> IdentifyGroup(A);
<12, 4>
> IdentifyGroup(G);
<12, 4>

Second, even his Theorem A states that $\mathrm{Aut}(D_{12})\cong \mathrm{Hol}(Z_6)$, and this is dihedral of order $12$.
