A response to this comment:
I understand that there are workarounds. I am wondering whether I am misinterpreting the manual.
Yes. The use of the word "function" in the manual may be confusing here. It is symbols that have attributes, not "functions". Attributes have an effect only at certain specific steps of the evaluation sequence, and only when the relevant symbol is the head of the expression that is being evaluated.
Here's the description of the evaluation sequence:
Given an expression
h[e1, e2, ...],
If the expression is a raw object (e.g., Integer, String, etc.), leave it unchanged.
Evaluate the head
h of the expression.
Evaluate each element $e_i$ of the expression in turn. If
h is a symbol with attributes
HoldAllComplete, then skip evaluation of certain elements.
(... several steps we don't care about now ...)
Use any built-in transformation rules for
h[e1, e2, ...] or for
What happens to your expression
h[2*3] is this:
h gets evaluated to
Function[g @@ Unevaluated[#]]. Now
Function[...] would get evaluated further if it had any applicable transformation rules, but it doesn't. Next the system looks at transformation rules for
Function[...][...] instead (notice the second set of brackets). There are rules for this, so the evaluation proceeds.
Notice that we never reached step 2. which would check any attributes of
h already got replaced at step 1. Thus the attributes of
h do not matter.
Another way to put it is that attributes do not belong to "functions". They belong to symbols, in your case to the symbol
Mathematica doesn't even have functions in the traditional sense. It has transformation rules.
h has an associated transformation rule which causes it to be replaced before any of its attributes could come into play.
Finally I'll just repeat J.M.'s comment. Use this:
h = Function[Null, g @@ Unevaluated[#], HoldAll]
I'll also note that
Function is a very special symbol that has very special evaluation rules: the evaluator will hold
y in an expression of the form
f[x][y] only when
Function (and in no other situation). In particular, it is not possible to bestow a custom symbol with this property.