In a discussion at the nForum:
https://nforum.ncatlab.org/comments.php?DiscussionID=7095
the following question arose.
Define an "oo-group" G to be a pointed connected type denoted BG.
(That is, when we regard a pointed connected type as an oo-group, we
write G for the oo-group and BG for its underlying type.) An "action"
of an oo-group G is a type family X : BG -> Type; it is said to be an
action "on" the type X(*), where *:BG is the basepoint. The action is
"regular" if \sum_{p:BG} X(p) is contractible.
Given a G-action X, define Aut_G(X) to be the oo-group with
BAut_G(X) = \sum_{P:BG->Type} || P=X ||.
and basepoint (X,refl). Then we have a BAut_G(X)-action defined by
Y : BAut_G(X) -> Type
Y(P,p) = P(*)
and because Y(X,refl) = X(*), this is an action on the same underlying type.
CONJECTURE: if the G-action X is regular, then so is the Aut_G(X)-action Y.
Any takers? This might be easy, or it might be hard; I haven't had
time to really think about it. When G is a set-group (i.e. BG is a
1-type) and X is a set, it is a classical fact; a proof is at
https://ncatlab.org/nlab/show/regular%20action. In that case there is
also a partial converse that if G acts transitively and faithfully on
X and Aut_G(X) acts regularly, then G also acts regularly. It's less
obvious how to define "transitive and faithful" in the oo-setting, but
that would be a natural next question.
Mike