Closure of regular languages under interchanging two different letters
Suppose that you guess that $\sigma$ and $\tau$ should be exchanged, where $\sigma$ appears before $\tau$ (in the original word).
You start reading the input word, letter by letter. When you encounter $\tau$, you can optionally act as if you read $\sigma$. Later on, when you encounter $\sigma$, you can optionally act as if you read $\tau$. From that point on, you accept the word if the original DFA accepted it.
Here is how to implement this in an NFA. Let the original DFA have states $Q$, initial state $q_0$, final states $F$, and transition function $\delta$. We construct a new NFA whose states are $\Sigma \times \Sigma \times Q \times \{0,1,2\}$. The initial states are $\{(\sigma,\tau,q_0,0) : \sigma \neq \tau\}$. The final states are $\Sigma \times \Sigma \times F \times \{2\}$. The most complicated bit is the transition function $\delta'$:

For $\kappa \neq \tau$, $\delta'((\sigma,\tau,q,0),\kappa) = \{ (\sigma,\tau,\delta(q,\kappa),0) \}$.
$\delta'((\sigma,\tau,q,0),\tau) = \{ (\sigma,\tau,\delta(q,\tau),0), (\sigma,\tau,\delta(q,\sigma),1) \}$.
For $\kappa \neq \sigma$, $\delta'((\sigma,\tau,q,1),\kappa) = \{ (\sigma,\tau,\delta(q,\kappa),1) \}$.
$\delta'((\sigma,\tau,q,1),\sigma) = \{ (\sigma,\tau,\delta(q,\sigma),1), (\sigma,\tau,\delta(q,\tau),2) \}$.
$\delta'((\sigma,\tau,q,2),\kappa) = \{ (\sigma,\tau,\delta(q,\kappa),2) \}$.