 Solution

From $ba{b}^{-1}=a^{-1}$ we obtain $ba=a^{-1}b$. Moving every occurrence of $b$ to the right and replacing $b^2$ by $a^n$ shows that every group element has the formThe element $b$ cannot lie in $⟨a⟩$: otherwise it would commute with $a$, forcing $a=a^{-1}$, contrary to $a$ having order $2n>2$. Thus the two cosets $⟨a⟩$ and $⟨a⟩b$ are disjoint and each has $2n$ elements. Hence every $n$-dicyclic group has order $4n$.

Existence is explicit. Put $\zeta =e^{\pi i/n}$ and takeThen $a$ has order $2n$,The $2n$ diagonal matrices $a^k$ and the $2n$ off-diagonal matrices $a^{k}b$ are distinct, so they form a dicyclic group of order $4n$.