Profinite closure of characteristic subgroup

Let $F$ be a free group of finite rank, and $K\subset F$ a finite index characteristic subgroup.

Let $\hat{F}$ be the profinite completion of $F$ (i.e. a free profinite group of same rank), and $\bar{K}$ the closure of $K$ in $\hat{F}$.

What are necessary and/or sufficient conditions for $\bar{K}$ to be characteristic in $\hat{F}$?

Is there an example of such $K$ for which $\bar{K}$ is not characteristic?

Motivation: I'm trying to answer the question whether there exists a finite (non-abelian) simple characteristic quotient of $F$. This will be possible only with a characteristic subgroup whose profinite closure is not characteristic.

This question was asked here in a more general setting. A sufficient condition was given in the answer, namely that $K$ be "hyper-characteristic" (or "isomorph-free"). However this doesn't help me, as the quotient by such a subgroup cannot be a non-abeian simple group, so I'm looking for other conditions.

• "(i.e., a free profinite group of the same rank)" The profinite completion is not just a profinite group, it's a profinite group along with a homomorphism from the original group. – YCor Mar 15 '18 at 10:45
• The first question is precisely a duplicate of the linked question (mathoverflow.net/questions/250809) by @rtz. It's a pity that rtz accepted his/her own answer, because it would be better gather new answers there, and because it would be worth isolating the second question (about existence of an example). – YCor Mar 15 '18 at 11:10
• @YCor The linked question considered general $F$, I was hoping that restricting to free $F$ might help to gather new answers. – ChanaG Mar 15 '18 at 12:10
• The other point is that finding an example can be quite energy-consuming (unless there's a pointer to the literature, or a simple example I would fail to see), so it would be more motivating if it's part of a single question. – YCor Mar 15 '18 at 12:24
• Here is something in answe to myself. I'm not 100% sure it's correct, and I don't know yet if it can be used to find concrete examples. Let $K_{n}$ be the intersection of all normal subgroups of $F$ of index $n$. Then $\hat{F}$ is the inverse limit of $F/K_{n}$, and the automorphism group of $\hat{F}$ is the inverse limit of automorphism groups of these quotients. Hence thw closure of $H\subset F$ is characteristic in $\hat{F}$ iff the image of $H$ is characteristic in $F/K_{n}$ for every $n$, – ChanaG May 23 '18 at 9:00