Wait, this isn't that hard. Let a positive integer $\lambda\ll 1$ be given. Let $A=[0,1/2n]\subset \mathbb{R}/\mathbb{Z}$. Let $\phi$ be the multiplication-by-$\lambda $ map; then $\phi^{-k}(A)$ is a union of $\lambda^k$ intervals with total measure $1/2n$. We let $B$ be the union of the sets $\phi^{-k}(A)$ for $k$ going from $0$ to $n-1$; we show that there isn't too much overlap, so that $1\ll |B|\leq 1/2$. Then the boundary of $B$ under $x\mapsto \lambda x$ is of size $O(1/2n)$.

Now let $f$ be the natural homomorphism of abelian groups $f:\mathbb{Z}/p\mathbb{Z}\to \mathbb{R}/\mathbb{Z}$. Of course, multiplication by $\lambda$ (i.e., addition $k$ times) gets taken to multiplication by $\lambda$. Because $B$ is the union of $O_\lambda(1)$ intervals,

• we have $p\ll_\lambda |f^{-1}(B)|\leq p/2 + O_\lambda(1)$,
• the boundary of $B$ under $x\mapsto \lambda x$ is of size $O(1/2n)+O_\lambda(1)$,
• the boundary of $B$ under $x\mapsto x+1$ is of size $O_\lambda(1)$,

and so the problem is solved.

(Note: this is very close to what Fiz does.)

Wait, this isn't that hard. Let a positive integer $\lambda\ll 1$ be given. Let $A=[0,1/2n]\subset \mathbb{R}/\mathbb{Z}$. Let $\phi$ be the multiplication-by-$\lambda $ map; then $\phi^{-k}(A)$ is a union of $\lambda^k$ intervals with total measure $1/2n$. We let $B$ be the union of the sets $\phi^{-k}(A)$ for $k$ going from $0$ to $n-1$; we show that there isn't too much overlap, so that $1\ll |B|\leq 1/2$. Then the boundary of $B$ under $x\mapsto \lambda x$ is of size $O(1/2n)$.

Now let $f$ be the natural homomorphism of abelian groups $f:\mathbb{Z}/p\mathbb{Z}\to \mathbb{R}/\mathbb{Z}$. Of course, multiplication by $\lambda$ (i.e., addition $\lambda$ times) gets taken to multiplication by $\lambda$. Because $B$ is the union of $O_{\lambda,n}(1)$ intervals,

• we have $p\ll_\lambda |f^{-1}(B)|\leq p/2 + O_{\lambda,n}(1)$,
• the boundary of $B$ under $x\mapsto \lambda x$ is of size $O(1/2n)+O_\lambda(1)$,
• the boundary of $B$ under $x\mapsto x+1$ is of size $O_\lambda(1)$,

and so the problem is solved.

(Note: this is very close to what Fiz does.)

Wait, this isn't that hard. Let a positive integer $\lambda\ll 1$ be given. Let $A=[0,1/2n]\subset \mathbb{R}/\mathbb{Z}$. Let $\phi$ be the multiplication-by-$2$ map; then $\phi^{-k}(A)$ is a union of $\lambda^k$ intervals with total measure $1/2n$. We let $B$ be the union of the sets $\phi^{-k}(A)$ for $k$ going from $0$ to $n-1$; we show that there isn't too much overlap, so that $1\ll |B|\leq 1/2$. Then the boundary of $B$ under $x\mapsto 2x$ is of size $O(1/2n)$.

Now let $f$ be the natural homomorphism of abelian groups $f:\mathbb{Z}/p\mathbb{Z}\to \mathbb{R}/\mathbb{Z}$. Of course, multiplication by $\lambda$ (i.e., addition $k$ times) gets taken to multiplication by $\lambda$. Because $B$ is the union of $O_\lambda(1)$ intervals,

• we have $p\ll_\lambda |f^{-1}(B)|\leq p/2 + O_\lambda(1)$,
• the boundary of $B$ under $x\mapsto 2x$ is of size $O(1/2n)+O_\lambda(1)$,
• the boundary of $B$ under $x\mapsto x+1$ is of size $O_\lambda(1)$,

and so the problem is solved.

(Note: this is very close to what Fiz does.)

Wait, this isn't that hard. Let a positive integer $\lambda\ll 1$ be given. Let $A=[0,1/2n]\subset \mathbb{R}/\mathbb{Z}$. Let $\phi$ be the multiplication-by-$\lambda $ map; then $\phi^{-k}(A)$ is a union of $\lambda^k$ intervals with total measure $1/2n$. We let $B$ be the union of the sets $\phi^{-k}(A)$ for $k$ going from $0$ to $n-1$; we show that there isn't too much overlap, so that $1\ll |B|\leq 1/2$. Then the boundary of $B$ under $x\mapsto \lambda x$ is of size $O(1/2n)$.

Now let $f$ be the natural homomorphism of abelian groups $f:\mathbb{Z}/p\mathbb{Z}\to \mathbb{R}/\mathbb{Z}$. Of course, multiplication by $\lambda$ (i.e., addition $k$ times) gets taken to multiplication by $\lambda$. Because $B$ is the union of $O_\lambda(1)$ intervals,

• we have $p\ll_\lambda |f^{-1}(B)|\leq p/2 + O_\lambda(1)$,
• the boundary of $B$ under $x\mapsto \lambda x$ is of size $O(1/2n)+O_\lambda(1)$,
• the boundary of $B$ under $x\mapsto x+1$ is of size $O_\lambda(1)$,

and so the problem is solved.

(Note: this is very close to what Fiz does.)