Let $\tau$ denote the permutation matrix corresponding to $(1,2,\ldots,n)$. Consider it first as an element of $\mathrm{M}_n(q^2)$.
This matrix has minimal polynomial equal to $X^n-1$, which is equal to its characteristic polynomial. It is therefore cyclic, and its centralizer is isomorphic to $\mathbb{F}_{q^2}$-algebra $\mathbb{F}_{q^2}[X]/(X^n-1)\simeq \mathbb{F}_{q^{2n}}$. Mapping $X\to \bar{X}^t$ defines a field automorphism of order $2$ of $\mathbb{F}_{q^{2n}}$ which, by Hilbert 90 (which is an overkill, but does the job), restricts to a surjective map $\mathbb{F}_{q^{2n}}^\times\to \mathbb{F}_{q^n}^\times$. The centralizer you're seeking is precisely the kernel of this map, and is of carinality $$ \frac{q^{2n}-1}{q^n-1}=q^n+1.$$
Does this make sense?