This is exercise 3.5.9 in Kowalski.
Given a simple graph $G$, the Markov averaging operator $M$ on $G$ is the adjacency matrix, normalized such that each row sums to $1$.
The normalized spectral gap $\lambda_1(G)$ is the smallest non-zero eigenvalue of $I-M$.
Let $G_n$ be the Cayley graph of $S_n$, with the generators being all transpositions. The problem is to show that $\lambda_1(G_n)=2/n$.
However, this doesn't appear to be true even for $n=3$. For $G_3$, $$M=\frac{1}{3}\begin{pmatrix}0&1&1&1&0&0\\1&0&0&0&1&1\\1&0&0&0&1&1\\1&0&0&0&1&1\\0&1&1&1&0&0\\0&1&1&1&0&0\\\end{pmatrix}$$
The eigenvalues of $I-M$ are $0,1,2$ ($1$ has multiplicity four). The smallest non-zero eigenvalue isn't $2/3$, as expected.
My guess is that I'm misunderstanding the definition of $M$, which is definition 3.2.15 in Kowalski.