TITLE: Centralizer of $Inn(G)$ in $Aut(G)$
QUESTION [5 upvotes]: Can the centralizer of Inn(G) in Aut(G), where G is preferably any non-abelian finite one, equal to Inn(G) itself? Clearly, such centralizer contains all $f$ in Aut(G) where $f(g)g^{-1}$ are in Z(G).

REPLY [4 votes]: I was fiddling with this, but never had time to finish.  Maybe it is interesting to someone.

We define two bijections that help to describe the centralizer of the inner automorphism group in the full automorphism group.

exp : Hom(G,Z(G)) → CAut(G)(Inn(G)) : ζ ↦ ( g ↦ gζ(g) )
log : CAut(G)(Inn(G)) → Hom(G,Z(G)) : φ ↦ ( g ↦ g−1 φ(g) )

We need to show these are well-defined (that is they have the specified ranges).
Proof: If ζ is a homomorphism from G to Z(G), then exp(ζ) : G → G : g ↦ gζ(g) is indeed an automorphism, with inverse exp(−ζ) defined by −ζ, the homomorphism from G to Z(G) that takes g to ζ(g)−1.  The inner automorphism defined by h takes g to h−1gh = gh.  If one applies exp(ζ) and then conjugation by h, one gets ghζ(g), since ζ(g) is central and so ζ(g)h = ζ(g).  On the other hand, if one conjugates by h and then applies exp(ζ), then one gets ghζ(gh), but since ζ is a homomorphism into an abelian group, ζ is constant on conjugacy classes, and ghζ(gh) = ghζ(g).  Hence every exp(ζ) is in CAut(G)(Inn(G)).  Conversely, if φ is an automorphism of G commuting with all inner automorphisms, then φ(x)φ(h) = φ(xh) = φ(x)h, so φ(h) and h define the same automorphism of G, and so φ(h) and h differ by some element of the center of G.  Define log(φ):G→Z(G) implicitly from φ(h) = h log(φ)(h).  It is not hard to check that log(φ) is a homomorphism from G to Z(G).
Notice that log(exp(ζ)) = ζ and exp(log(φ))=φ, so these are mutually inverse bijections.

Suppose then that Inn(G) = CAut(G)(Inn(G)).  This means that exp(Hom(G,Z(G))) = Inn(G).  This means that log : Inn(G) → Hom(G,Z(G)) is bijective!  However, log(∧h) = ( g ↦ [g,h] ) has a very simple form for the inner automorphism ∧h defined by conjugation by h.  In particular, [G,G] ≤ Z(G) for log(∧h) to even make sense, and so G has nilpotency class at most 2.  Notice that the kernel of log(∧h) contains all of Z(G), for any h.  Indeed, the kernel of log(∧h) is CG(h), and so all centralizers are normal, which is weird.
Another sick thing: log(∧hk)(g) = [g,hk] = [g,k][g,h] = [g,h][g,k] = (log(∧h) + log(∧k))(g), so in fact log is an isomorphism of abelian groups G/Z(G) ≅ Hom(G,Z(G)).
Unfortunately, I cannot find a good element of Hom(G,Z(G)).  The transfer is useless, and the central series induction arguments give inequalities pointing the wrong way.
For finite groups, presumably this is taken care of by counting as in theorem 1 of:

Sanders, P. R.
  "The central automorphisms of a finite group."
  J. London Math. Soc. 44 (1969) 225–228.
  MR248208
  DOI:10.1112/jlms/s1-44.1.225

but it didn't seem to address infinite (nilpotent) groups.  The way it is done in this (very nice) paper mentioned by Steve certainly doesn't work out for infinite groups.  In particular, Lemma I would be wonderful to have more concretely with homomorphisms rather than dualizing finite abelian groups a zillion times.

Curran, M. J.; McCaughan, D. J.
  "Central automorphisms that are almost inner."
  Comm. Algebra 29 (2001), no. 5, 2081–2087.
  MR1837963
  DOI:10.1081/AGB-100002170