ls Ordering[Ordering[list]] optimal?

Question

Given a list list with unique elements, the task is to replace each element by its position in Sort[list]. For example,

list = {"A", "B", "D", "C", "Z", "W"};
Position[Sort[list], #][[1, 1]] & /@ list

{1, 2, 4, 3, 6, 5}

Much more efficient is to call Ordering twice:

Ordering[Ordering[list]]

{1, 2, 4, 3, 6, 5}

When applied on a permutation of Range[length] this operation does nothing:

list = {2, 10, 1, 4, 8, 6, 3, 9, 5, 7};
Ordering[Ordering[list]]

{2, 10, 1, 4, 8, 6, 3, 9, 5, 7}

Question: is there a more efficient way of doing this operation, making a single function call instead of calling Ordering twice?

---

benchmarks

Solutions are given from fastest to slowest:

L = RandomReal[{0, 1}, 10^7];

(* J.M.'s undocumented InversePermutation usage *)
R0 = InversePermutation[Ordering[L]]; // AbsoluteTiming // First
(* 2.39154 *)

(* Henrik Schumacher *)
R1[[Ordering[L]]] = R1 = Range[Length[L]]; //AbsoluteTiming//First
(* 2.42264 *)

(* original post *)
R2 = Ordering[Ordering[L]]; //AbsoluteTiming//First
(* 4.20186 *)

(* J.M. *)
R3 = PermutationList[InversePermutation[FindPermutation[L]]]; //AbsoluteTiming//First
(* 4.74717 *)

(* check *)
R0 == R1 == R2 == R3
(* True *)

Answer 1

No, Ordering[Ordering[list]] not optimal. And yes, there is a faster method:

list = RandomReal[{-1, 1}, 1000000];

First@RepeatedTiming[
  a[[Ordering[list]]] = a = Range[Length[list]];
  ]

First@RepeatedTiming[
  b = Ordering[Ordering[list]]
  ]

a == b

0.13

0.236

True

Edit

J.M.'s second suggestion is more concise and at least as fast if not slightly faster:

c = InversePermutation[Ordering[list]]; // RepeatedTiming // First

c == b

0.124

True