= Solution
A relation $r$ on a set $x$ is <well-founded relation>[well-founded] when every nonempty subset of $x$ has an $r$-minimal element. It is <extensional relation>[extensional] when distinct elements have distinct predecessor sets:
$$
a\ne b
\Longrightarrow
\{z\in x:z\,r\,a\}
\ne
\{z\in x:z\,r\,b\}.
$$
The <Mostowski collapse theorem> states that every well-founded extensional relation on a set is uniquely isomorphic to membership on a transitive set; the collapse is recursively
$$
\pi(a)=\{\pi(b):b\,r\,a\}.
$$
For the requested example, let $A=V_{\omega+1}$ and define
$$
x_0=A,
\qquad
x_n=A\cup\{x_0,\ldots,x_{n-1}\}\quad(n\geq1),
\qquad
x=\{x_n:n\in\omega\}.
$$
No $x_n$ belongs to $A$, and therefore
$$
x_m\in x_n\quad\Longleftrightarrow\quad m<n.
$$
The map $x_n\mapsto n$ is an isomorphism from membership on $x$ to membership on the von Neumann ordinal $\omega$, so the Mostowski collapse is $\omega$. On the other hand, $x_0=V_{\omega+1}$ already has rank $\omega+1$, and hence the rank of $x$ is greater than $\omega$.
For statement (i), suppose $(x,r)$ is isomorphic to $(y,\in)$ with $y$ transitive. Membership is well-founded by the axiom of foundation. It is extensional on $y$ because transitivity ensures that the predecessors in $y$ of an element $a\in y$ are exactly the elements of $a$, and the axiom of extensionality distinguishes different sets. Both properties are preserved by isomorphism. Thus (i) is always true.
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