How does one find the rank of a conjugate closure of a subset?

In particular, I am studying this group: Let $K_n$ be the group with $n$ generators $x_1,\cdots,x_n$ satisfying the relation that each $x_i$ commutes with all its conjugates, where $i=1,\cdots,n$. Thus there is a group homomorphism $f_{n}:K_n\to K_{n-1}$ mapping $x_i$ to $x_i$ when $i\leq n$ and maps $x_n$ to the identity element. Consider the kernel of $f$: it is the conjugate closure of the one-element subset $\{x_n\}$ of $K_n$. Then what is the rank (I mean the minimum number of generators) of the conjugate closure of $\{x_n\}$?

How does one find the rank of a normal closure of a subset?

In particular, I am studying this group: Let $K_n$ be the group with $n$ generators $x_1,\cdots,x_n$ satisfying the relation that each $x_i$ commutes with all its conjugates, where $i=1,\cdots,n$. Thus there is a group homomorphism $f_{n}:K_n\to K_{n-1}$ mapping $x_i$ to $x_i$ when $i\leq n$ and maps $x_n$ to the identity element. Consider the kernel of $f$: it is the conjugate closure of the one-element subset $\{x_n\}$ of $K_n$. Then what is the rank (I mean the minimum number of generators) of the normal closure of $\{x_n\}$?