Are the separability and autoseparability equivalent for (locally) compact topological group?

Question

Definition. A topological group $G$ is called autoseparable if there exists a countable subset $S\subset G$ and a sequence $(f_n)_{n\in\omega}$ of automorphisms of $G$ such that for any neighborhood $U\subseteq G$ of the unit of $X$ we have $X=\bigcup_{n=0}^\infty f_n(US)$.

It is clear that each separable topological group is autoseparable. The converse is not true as shown by the following

Example. Each topological vector space $X$ is autoseparable, which is witnessed by the set $S=\{0\}$ and the sequence $(f_n)_{n\in\omega}$ of automorphisms $f_n(x)=nx$.

Problem 1. Are the separability and autoseparability equivalent for (locally) compact topological groups?

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Added in Edit. In can be shown that for any finite field $\mathbb F$ and any cardinal $\kappa$ the Tychonoff power $\mathbb F^\kappa$ is autoseparable, which means that the answer to the above problem is negative. So, Problem 1 transforms into

Problem 2. Which compact topological groups are autoseparable?

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Added in the next Edit. The answer to this MO-question implies that for any infinite cardinal $\kappa$ the compact topological group $\mathbb T^\kappa$ is autoseparable, which implies that each ($\sigma$-compact locally) compact abelian topological group embeds into an autoseparable (locally) compact abelian topological group.

Answer 1

Extended comment. Let me first rephrase the question. Say that a topological group $G$, a subgroup $A$ of $\mathrm{Aut}(G)$, and a subset $S$ of $G$ satisfy: for every neighborhood $U$ of $1$ in $G$ we have $G=\bigcup_{f\in A,x\in S}f(Ux)$. Say that $G$ has Property (V) if it has such $A$, $S$ countable, and (V') is it has such $A$ countable with $S=\{1\}$.

You're asking when $G$ locally compact has Property (V). It clearly implies $\sigma$-compact and is implied by separable, and more specifically the question ask about equivalence with being separable.

Let me do two reductions.

(1) If there's a non-separable locally compact abelian group $G$ satisfying (V), then there another one (a quotient of $G$) satisfying (V').

Indeed, the quotient by the closure of the subgroup generated by $\bigcup_{f\in A}f(S)$ works.

(2) If there's a non-separable locally compact group $G$ satisfying (V'), then there is a compact one (a compact normal subgroup of $G$).

Proof: since $G$ is $\sigma$-compact, there is a compact normal subgroup $K$ of $G$ such that $G/K$ is second-countable. Then $W=\bigcap_{f\in A}f(K)$ is such that $G/W$ is second-countable, and in addition is $A$-invariant. Hence if $G$ is not separable, then $W$ is not separable. Also $W$ satisfies (V'): indeed let $U$ be a neighborhood of $1$ in $W$, then there exists a neighborhood $U'$ of $1$ in $G$ such that $U=U'\cap W$. Then $G=\bigcup_{f\in A}f(U)$. Intersecting with $W$ this yields $W=\bigcup_{f\in A}f(U')$.

The latter proof shows:

If $G$ satisfies (V') for some $A$ and $W$ is an $A$-invariant closed normal subgroup, then $W$ also satisfies (V').

Combining all this, if there is a non-separable locally compact abelian group with (V'), then there is one that in addition is compact, and either connected or profinite.