How do I make Conjugate behave more consistently?

list = {a b, 2 a b, a (b + c), a (b + 2 c), a (b + c (d + e))};

If you never want Conjugate to distribute, use Inactive

Inactive[Conjugate] /@ list

If you want Conjugate to always distribute

conj[expr_] := 
 ComplexExpand[expr, Variables@Level[expr, {-1}], 
   TargetFunctions -> Conjugate] // Simplify

conj@*Conjugate /@ list

(*  {Conjugate[a] Conjugate[b], 2 Conjugate[a] Conjugate[b], 
 Conjugate[a] (Conjugate[b] + Conjugate[c]), 
 Conjugate[a] (Conjugate[b] + 2 Conjugate[c]), 
 Conjugate[a] (Conjugate[b] + Conjugate[c] (Conjugate[d] + Conjugate[e]))}  *)

If any variables are Reals, say a and c, then use Simplify

Simplify[%, Element[{a, c}, Reals]]

(*  {a Conjugate[b], 2 a Conjugate[b], a (c + Conjugate[b]), 
 a (2 c + Conjugate[b]), a (Conjugate[b] + c (Conjugate[d] + Conjugate[e]))}  *)

Mathematica does not know the nature of your symbols. It knows for sure that 2 is a real number - so it can use proper distribution. You can always specify your variables to get a more consistent answer. For example,

Simplify[Conjugate[a (b + c)], Assumptions -> {a, b, c} ∈ Reals]

a (b + c)

Simplify[Conjugate[a (b + c)], Assumptions -> {{a, b} ∈ Reals}]

a (b + Conjugate[c])

Simplify[Conjugate[a (b + c)], Assumptions -> {{a} ∈ Reals}]

a Conjugate[b + c]

Or you can use Expand as well to see all the terms distinctly

Expand[Simplify[Conjugate[a (b + c)], Assumptions -> {a, b, c} ∈ Reals]]

a b + a c

And last but not the least, the most general case

Simplify[Conjugate[a (b + c)], Assumptions -> {a, b, c} ∈ Complexes]

Conjugate[a (b + c)]