Here is a proof of Gordon's claim. We will prove something slightly stronger.

Claim. Let $G$ be a $d$-regular graph with $d$ odd. Then for every $e \in E(G)$, there is an even number of Hamiltonian cycles using $e$.

Proof. Let $e=uv$ and let $\mathcal{H}$ be the set of Hamiltonian paths in $G$ that start at $u$ and use the edge $e$. Suppose that $P$ is such a path with ends $u$ and $w$. Note that for any edge $wv$ with $v \neq u$, there is exactly one other Hamiltonian path $P_{wv} \in \mathcal{H}$ contained in $P \cup \{wv\}$. Create an auxiliary graph $G'$ with $V(G')=\mathcal{H}$ and if $P \in \mathcal{H}$ has ends $uw$, then $P$ is adjacent to $P_{wv}$ for all $v \neq u$. Finish by observing that there is a 1-1 correspondence between Hamiltonian cycles in $G$ using $e$ and odd-degree vertices of $G'$. Hence, there are an even number of them, as required.

Since your graph has at least one Hamiltonian cycle, it necessarily has at least two of them by the claim.

Here is a proof of Gordon's claim. We will prove something slightly stronger.

Claim. Let $G$ be a $d$-regular graph with $d$ odd. Then for every $e \in E(G)$, there is an even number of Hamiltonian cycles using $e$.

Proof. Let $e=uv$ and let $\mathcal{H}$ be the set of Hamiltonian paths in $G$ that start at $u$ and use the edge $e$. Suppose that $P$ is such a path with ends $u$ and $w$. Note that for any edge $wv$ with $v \neq u$, there is exactly one other Hamiltonian path $P_{wv} \in \mathcal{H}$ contained in $P \cup \{wv\}$. Create an auxiliary graph $G'$ with $V(G')=\mathcal{H}$ and if $P \in \mathcal{H}$ has ends $uw$, then $P$ is adjacent to $P_{wv}$ for all edges $wv$ with $v \neq u$. Finish by observing that there is a 1-1 correspondence between Hamiltonian cycles in $G$ using $e$ and odd-degree vertices of $G'$. Hence, there are an even number of them, as required.

Since your graph has at least one Hamiltonian cycle, it necessarily has at least two of them by the claim.