= Solution
From $bab^{-1}=a^{-1}$ 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 form
$$
a^k\quad\hbox{or}\quad a^kb,
\qquad 0\leq k<2n.
$$
The element $b$ cannot lie in $\langle a\rangle$: otherwise it would commute with $a$, forcing $a=a^{-1}$, contrary to $a$ having order $2n>2$. Thus the two cosets $\langle a\rangle$ and $\langle a\rangle b$ 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 take
$$
a=\begin{pmatrix}\zeta&0\\0&\zeta^{-1}\end{pmatrix},
\qquad
b=\begin{pmatrix}0&1\\-1&0\end{pmatrix}.
$$
Then $a$ has order $2n$,
$$
b^2=-I=a^n,
\qquad bab^{-1}=a^{-1}.
$$
The $2n$ diagonal <matrices> $a^k$ and the $2n$ off-diagonal <matrices> $a^kb$ are distinct, so they form a <dicyclic group> of order $4n$.
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