Simplifying simple signed expression ( such as $-x(x-1) \to x(1-x)$ ) based on assumptions

You can use ComplexityFunction option to "penalize" the number of terms with a negative sign:

cf = (LeafCount[#] + 100 Count[#, _?Internal`SyntacticNegativeQ, ∞] &); 

Simplify[(-2*x^2*(-1 + x^2))/135, ComplexityFunction -> cf]

2/135 x^2 (1 - x^2)

As chris noted in comments, this gives an unsatisfactory result for the following example:

Simplify[(Sqrt[1 - x] - 3*(-1 + x)*ArcTan[3*Sqrt[1 - x]])/(4*Pi), 
 ComplexityFunction -> cf]

(Sqrt[1 - x] + (3 - 3 x) ArcTan[3 Sqrt[1 - x]])/(4 π)

since it expands - 3*(-1 + x) to (3-3x).

A simple modification of cf to cf2 normalizing the second term by the LeafCount of input expression:

cf2 = LeafCount[#] + 100 Count[#, _?Internal`SyntacticNegativeQ, ∞]/ LeafCount[#] &; 

gives the desired result for both examples:

Simplify[(-2*x^2*(-1 + x^2))/135, ComplexityFunction -> cf2]

2/135 x^2 (1 - x^2)

Simplify[(Sqrt[1 - x] - 3*(-1 + x)*ArcTan[3*Sqrt[1 - x]])/(4*Pi), 
 ComplexityFunction -> cf2]

(Sqrt[1 - x] + 3 (1 - x) ArcTan[3 Sqrt[1 - x]])/(4 π)

Needless to say, I am sure there will be examples where both will fail to give the desired result.

Second Answer

Sticking with the problem of having a product's syntactic sign agree with the sign of its real value under $Assumptions, assuming it's real-valued (and not the problem of just supressing an initial minus sign regardless of the sign of the expression)...

The idea of syntacticallyCorrectSignForm is to convert any negative factors to positive factors and and keep track of the sign changes:

syntacticallyCorrectSignForm[e_Times] := Module[{sign = 1, res},
  res = If[
      TrueQ@Simplify[# < 0],
      sign = -sign
      ; If[Internal`SyntacticNegativeQ[#], Defer[#], #] &@Simplify[-#],
      #] & /@ e;
  sign*res]

OP's example (from comment below):

Assuming[x < 1 && x > 0 && b > 0,
 syntacticallyCorrectSignForm[b (-2*x^2*(-1 + x^2))/135]
 ]

Powers of negative expressions are difficult, since (-y)^3 automatically evaluates to -(y^3). The use of Defer in syntacticallyCorrectSignForm allows the output expression to be copied as input.

Assuming[x < 1 && x > 0 && b > 0 && y < 0,
 syntacticallyCorrectSignForm[b (-2*x^2*y^3*(-1 + x^2))/135]
 ]

---

Original Answer

Here's what I came up with, understanding the question to say that we want a negative sign when the expression is negative:

cf[e_Times] := 
  Boole[Internal`SyntacticNegativeQ[e] =!= 
    TrueQ@Simplify[
      Reduce[$Assumptions \[Implies] e < 0], $Assumptions]];
cf[e_Plus] := cf /@ e;
cf[e_] := 0;
xf = # /. e_Plus /; Simplify[
        Reduce[$Assumptions \[Implies] e < 0],
    $Assumptions
        ] :> -\[FormalX][-e] /. \[FormalX] -> Expand &;

OP's example:

Assuming[x < 1 && x > 0,
 Simplify[(-2*x^2*(-1 + x^2))/135, 
  ComplexityFunction ->
   (Simplify`SimplifyCount[#] + 10 cf[#] - 5 Boole[Head[#] === Times] &), 
  TransformationFunctions -> {Automatic, xf}
  ]
 ];
(*  2/135 x^2 (1 - x^2)  : This expression is positive under the assumptions *)

The negative of the OP's example:

Assuming[x < 1 && x > 0,
 Simplify[(2*x^2*(-1 + x^2))/135,
  ComplexityFunction ->
   (Simplify`SimplifyCount[#] + 10 cf[#] - 5 Boole[Head[#] === Times] &), 
  TransformationFunctions -> {Automatic, xf}
  ]
 ]
(*  -(2/135) x^2 (1 - x^2)  : This expression is negative under the assumptions *)

Another example from a comment:

Assuming[x < 1 && x > 0 && b > 0,
 Simplify[b (2*x^2*(-1 + x^2))/135, 
  ComplexityFunction ->
   (Simplify`SimplifyCount[#] + 10 cf[#] - 5 Boole[Head[#] === Times] &), 
  TransformationFunctions -> {Automatic, xf}
  ]
 ]
(*  -(2/135) b x^2 (1 - x^2)  *)

Note that there's nothing that can be done about expressions like -(1-x^2), which evaluates automatically to x^2 - 1, AFAICT.