least two Hamiltonian cycles $(m_1,\cdots, m_{n}, m_1)$ and $(m_1, m_n,\cdots, m_1)$ in $H_{G'}$. Define a map $\varphi$ from $H_G$ to $\mathcal
{P}(H_{G'})$ as follows: $$\varphi(\mathcal{H})= \{(m_1, \cdots, m_{n}, m_1), (m_1, m_n,
\cdots, m_1)\}.
$$ Let $Im\varphi$ denote the image set of the map $\varphi$, and let $\mathcal{H}'=m'_1\cdots m'_{n}m'_1$ be a different Hamiltonian cycle from $\mathcal{H}$ in $H_G$. Then $$\varphi(\mathcal{H}) \cap \varphi(\mathcal{H}')=\emptyset \qquad
\operatorname{and} \qquad \cup Im\varphi=H_{G'}.$$*\
**Proof.** Due to the symmetry of the digraph $G'$, and noting $n\geq 3$, for any Hamiltonian cycle $(m_1,\cdots, m_{n}, m_1)$ in $H_{G'}$, there must be a different Hamiltonian cycle $(m_1,
m_n,\cdots, m_1)$ in $H_{G'}$. These two Hamiltonian cycles obviously has a pre-imagine, the Hamiltonian cycle $m_1\cdots
m_{n}m_1$ in $H_{G}$. Note $(m_1,\cdots, m_{n}, m_1)$ is in $\varphi(m_1\cdots m_{n}m_1)$. Hence, $\cup Im\varphi\supseteq
H_{G'}.$ Obviously, $\cup Im\varphi\subseteq H_{G'}$. Therefore $$\cup Im\varphi=H_{G'}.$$ Suppose there are two different Hamiltonian cycles $\mathcal{H}=m_1\cdots m_{n}m_1$ and $\mathcal{H}'=m'_1\cdots
m'_{n}m'_1$ in $H_{G}$. Let $N_{\mathcal{H}}(m_i)$ denote two neighbor vertices of vertex $m_i$ in $\mathcal{H}$. $\mathcal{H}$ and $\mathcal{H}'$ are different if and only if there exits a vertex $\{m_i\}$=$\{m'_j\}$ such that $N_{\mathcal{H}}(m_i)\neq
N_{\mathcal{H}'}(m'_j)$. Hence $(m_1,\cdots, m_{n}, m_1)$ is different from $(m'_1, \cdots, m'_{n}, m'_1)$ and $(m'_1, m'_n,
\cdots, m'_{2}, m'_1)$, that is $(m_1,\cdots, m_{n},
m_1)\notin\varphi(\mathcal{H}')$. Similarly, 