This question is related to this problem on MO about generically Haar-null sets in locally compact Polish groups but is more concrete.
First we recall the definition of a generically Haar-null set in a Polish group.
Let $2^\omega=\{0,1\}^\omega$ be the Cantor cube and $\lambda$ be the standard product measure on $2^\omega$ (or any other atomless strictly positive probability Borel $\sigma$-additive measure on $2^\omega$).
Let $\mathcal N_\lambda$ be the $\sigma$-ideal of subsets of $\lambda$-measure zero in $2^\omega$.
A subset $A$ of a Polish group $X$ is called
$\bullet$ Haar-null if there exists a continuous map $f:2^\omega\to X$ such that $f^{-1}(A+x)\in\mathcal N_\lambda$ for all $x\in X$;
$\bullet$ generically Haar-null if the set $\{f\in C(2^\omega,X):\forall x\in X\;f^{-1}(A+x)\in\mathcal N_\lambda\}$ is comeager in the function space $C(2^\omega,X)$.
Here $C(2^\omega,X)$ is the (Polish) space of continuous maps from $2^\omega$ to $X$, endowed with the compact-open topology.
It can be shown that the above definitions of (generically) Haar-null set are equivalent to those given in this MO-post.
I cannot check if the following concrete Haar-null set in generically Haar-null.
In the compact topological group $X=\prod_{n\in\omega}\mathbb Z/2^n\mathbb Z$ consider the $F_\sigma$-subset $A=\{(x_n)_{n\in\omega}\in X:\exists m\in\omega\;\forall n\ge m\;x_n\notin 0+2^n\mathbb Z\}$ of Haar measure 1. Let also $\mathbf 1=(1+2^n\mathbb Z)_{n\in\omega}\in X$. It is easy to see that the set $B=(A+\mathbf 1)\setminus A$ has Haar measure zero and hence is Haar-null.
Question. Is $B$ generically Haar-null in the compact Polish group $X$?
