Translation invariant ultrafilters?

Question

Is there an ultrafilter $\mathcal{U}$ on $\mathbb{N}$ such that $\mathcal{U}-n = \mathcal{U}$ for all $n \in \mathbb{N}$? Here, $$\mathcal{U}-n = \{ A - n: A \in \mathcal{U} \}$$ and $$A - n = \{ m \in \mathbb{N}: m+n \in A \}$$

Answer 1

Let $E$ and $O$ be the set of even and odd naturals respectively, and note that $E-1=O$ and $O-1=E$. If $\mathcal{U}$ is an ultrafilter, either $E\in\mathcal{U}$ and $O\not\in\mathcal{U}$ or $E\not\in\mathcal{U}$ and $O\in\mathcal{U}$. Either way, there is an $X\in\mathcal{U}$ such that $X-1\not\in\mathcal{U}$, so $\mathcal{U}\not=\mathcal{U}-1$.

By a similar argument (consider "alternating blocks of size $n$") we get $\mathcal{U}\not=\mathcal{U}-n$ for every ultrafilter $\mathcal{U}$ and every nonzero $n\in\mathbb{N}$. So in fact we can never have the equality in the OP happen for even a single (nonzero) natural number.