DFT symmetry vs DFT duality in Richard Lyons' "Understanding DSP"
1. Duality between discrete frequency and discrete time domain.


DFT Duality is generally referred to the duality of DFT-IDFT pairs. This in turn comes from the similarity between analysis and synthesis expressions of DFT and IDFT. 
$$X[k] = \sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{N}nk}$$
$$x[n] = \frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j\frac{2\pi}{N}nk}$$
(Although I will encourage you to understand DFT as projection of Finite-length sequence x[n] onto Discrete Fourier Orthogonal Basis Vectors, and IDFT as representation of x[n] as linear combination of those orthogonal fourier basis vectors, where coefficients of linear combination is are the DFT coefficients $X[k]$. You might want to read this answer : DFT as Projection on Orthogonal Fourier Basis Vectors)

By Duality between the DFT-IDFT pairs I mean, for example, DFT of discrete $\delta[n]$ is a constant $1$ in freq-domain and DFT of constant $1$ will be discrete $\delta[k]$ in freq-domain. And other example would be DFT of discrete $rect$ sequence i.e. sequence of $M$ $1$'s in an $N&gt;M$ length sequence. It has its DFT as follows:
$$X[k] = e^{-j\frac{\pi}{N}(M-1)k}\frac{\sin[\frac{\pi}{N}Mk]}{\sin[\frac{\pi}{N}k]}, \forall \ k \in \ \{0,1,2,...,(N-1)\}$$
And, if we took DFT of $x[n] = e^{-j\frac{\pi}{N}(L-1)n}\frac{\sin[\frac{\pi}{N}Ln]}{\sin[\frac{\pi}{N}n]}$, we will get a sequence of $L$ consecutive $1$'s in an $N$ length DFT. We can just run the steps backwards to compute the DFT of this $x[n]$.

For continuous time fourier transform, duality would mean the following : if $f(t)$ has fourier transform $F(\Omega)$ then a time domain function $F(t)$ will have its fourier transform as $2\pi f(-\Omega)$.


  2. Symmetry in Discrete Fourier Transform Coefficients.


DFT Symmetry in the book is mentioned in context of symmetry properties in DFT coefficients when DFT of a real valued time domain sequence $x[n]$ is computed. That is DFT coefficients of all real valued $x[n]$ are conjugate symmetric modulo $N$.
$$X[k] = X^*[(N-k) \ \mod \ N]$$