Let $P$ be the transition matrix of a simple random walk on a cycle $\mathbb{Z}_n$. Let $f$ be a bijection, specifically a transposition, of $\mathbb{Z}_n$ i.e. $f(0)=1,f(1)=0$ and $f(j)=j \;\forall 1<j<n$. Let $\Pi$ be the permutation matrix defined by $f$ i.e. $\Pi = (\pi_{ij})_{i,j\in \mathbb{Z}_n}$ is s.t. $\pi_{ij}=1$ if $j=f(i)$ and $0$ otherwise.
I am trying to find the order of spectral gap of the Markov chain $\Pi P$?
I know that the spectral gap of Markov chain $P$ is $O(n^{-2})$ because:
We can think of the simple random walk on $\mathbb{Z}_n$ as a simple random walk on the multiplicative group $W_n=\{w,w^2,\cdots,w^{n-1},1\}$ where $w=e^{2\pi i/n}$ i.e. $W_n$ are the $n^{th}$ roots of unity.
For $0\leq j \leq n-1$, let $\phi_j(w^k):=w^{kj}$. Thus
$$P\phi_j(w^k)=\frac{\phi_j(w^{k-1})+\phi_j(w^{k+1})}{2} =\frac{w^{kj-j}+w^{kj+j}}{2} = w^{kj}\left( \frac{w^{-j}+w^{j}}{2} \right) = \phi_j(w^k)\left( \frac{w^{-j}+w^{j}}{2} \right)$$
Hence $\phi_j$ is an eigenfunction of $P$ with eigenvalue $\lambda_j = \left( \frac{w^{-j}+w^{j}}{2} \right)=\cos\left(\frac{2\pi j}{n} \right)$.
So $\lambda_2 = \cos\left(\frac{2\pi}{n} \right) = 1-\frac{4\pi^2}{2n^2}+O(n^{-4})$ and so the spectral gap is of order $O(n^{-2})$.
I cannot use the same function $\phi_j(w^k):=w^{kj}$ for the Markov chain $\Pi P$ because the above is true except for $k=0,1$ because:
For $k=0$,
$$\Pi P \phi_j(w^k) = \frac{\phi_j(w^{k})+\phi_j(w^{k+2})}{2} = \frac{w^{kj}+w^{kj+2j}}{2} = w^{kj}\left( \frac{1+w^{2j}}{2} \right) = \phi_j(w^k)\left( \frac{1+w^{2j}}{2} \right)$$
For $k=1$,
$$\Pi P \phi_j(w^k) = \frac{\phi_j(w^{k})+\phi_j(w^{k-2})}{2} = \frac{w^{kj}+w^{kj-2j}}{2} = w^{kj}\left( \frac{1-w^{2j}}{2} \right) = \phi_j(w^k)\left( \frac{1-w^{2j}}{2} \right)$$
and so $\phi_j(w^k)=w^{kj}$ is not an eigenfunction of $\Pi P$ and so I'm unable to find the spectral gap order. Uniform distribution is stationary for $\Pi P$ as it is stationary for $P$. Also, $\Pi P$ need not be symmetric and hence not reversible with respect to uniform distribution and therefore I cannot use a known bound for spectral gap for reversible chains: $\frac{\Phi^2}{2}\leq\gamma\leq2\Phi$ where $\Phi$ is the Cheeger constant. Any ideas on how to find the eigenfunctions of $\Pi P$? Thanks.