Self-similarity in ultrafilters over N First, some notation:


*

*Set variables, $X, Y$, range over sets of natural numbers, $\mathbb{N}={1,2,3,..}$.

*Square brackets represent sets of natural numbers based on a formula.
$$[φ(\mathbf{n})]=\{φ(n)|n∈N\}$$
So, $[2n]=\{2,4,6…\}$ and $[\mathbf{n}^2] = {1, 4, 9, ..}$

*$X_Y$ = the selection of $X$ based on $Y$:
$$X_Y = \{X_k|k\in Y\}$$ where $X_k$ is the $k$-th member of $X$ (in numeric order).
So, $X_{[2n]}$ consists of every second element of X and $X_{[n^2]} = \{X_1, X_4, X_9,…\}$
Now, the question: Is there a non-principal ultrafilter, $F$, over $\mathbb{N}$ such that:


*

*For each $k \in \mathbb{N}$, $[k\mathbf{n}]\in F$.

*If $X \in F$, then, for every $k$,
$[k\mathbf{n}]_X \in F$.

 A: If I correctly understand your notation, the set $[kn]_X$ in requirement 2 is just $\{kx:x\in X\}$, so you are asking for the ultrafilter $F$ to be invariant under the operation $\mathbb N\to\mathbb N$ of multiplication by $k$ (for all $k$).  There is no ultrafilter that accomplishes this, even for a single $k>1$.  The reason is the theorem that, if an ultrafilter is mapped to itself by an operation $f$ on its underlying set, then the ultrafilter must contain the set of fixed-points of $f$.  
For the particular operation you asked about, the (proof of) the general theorem can be specialized to the following.  Partition $\mathbb N$ into the set $X$ of those numbers whose binary expansion ends with an even number of 0's (i.e., odd numbers, numbers congruent to 4 mod 8, etc.) and the set $Y$ of numbers whose binary expansion end with an odd number of 0's (congruent to 2 mod 4, or to 8 mod 16, etc.) An ultrafilter $F$ must contain exactly one of these two sets.  But $[2n]_X=Y$ and $[2n]_Y=X$, so $F$ cannot satisfy your requirement 2.
