Given a set $E$, let $\mathfrak{S}_E$ be the group of permutations of $E$.
Definition.$\ \ $ Let $E$ be a finite set, $\zeta$ a permutation of $E$, and $\overline{\zeta}$ the subgroup of $\mathfrak{S}_E$ generated by $\zeta$. We say that $\zeta$ is a cycle if, under the operation of $\overline{\zeta}$ on $E$, there exists a unique orbit which is not reduced to a single element. This orbit is called the support of $\zeta$ and denoted by $\text{supp}(\zeta)$. The order of $\zeta$ is the cardinality of its support.
Here the operation of $\mathfrak{S}_E$ on $E$ is the identity mapping. Thus for $\sigma \in \mathfrak{S}_E$ and $x \in E$, we define $\sigma.x = \sigma(x)$.
More precisely (correct me if I'm wrong) :
$\zeta$ is a cycle if and only if there exists a unique $x \in E$ such that $$\text{Card}\Big(\overline{\zeta}.x\Big) > 1,$$ and then $\text{supp}(\zeta) = \overline{\zeta}.x$.
It is easy to show that $$\text{Card}\Big(\overline{\zeta}.x\Big) > 1 \iff \zeta(x) \ne x.$$ In fact, the condition is necessary, for if $\zeta(x) = x$, then $\zeta^n(x) = x$ for all $n \in \mathbb{Z}$, so $\overline{\zeta}.x = \{x\}$. Conversely, if $\zeta(x) \ne x$, then $\{x,\zeta(x)\} \subseteq \overline{\zeta}.x$.
The book claims that $y \in \text{supp}(\zeta)$ if and only if $\zeta(y) \ne y$. If everything above is true, then $\text{supp}(\zeta) = \{x\}$, where $x$ is the unique element of $E$ whose orbit is not reduced to a single element.
But this is not true, since a transposition is a cycle of order $2$. Therefore something above is incorrect.
Question.$\ \ $ What am I misunderstanding?
No, $x$ is not unique. What is unique is just the orbit $\overline \zeta\cdot x$, as a subset of $E$. If $y$ is another element of that orbit, we have $\overline \zeta\cdot x = \overline\zeta \cdot y$, yet $x\ne y$.