Find all real numbers $m$ such that $$\frac{1-m}{2m} \in \{x\ |\ m^2x^4+3mx^3+2x^2+x=1\ \forall \ x\in \mathbb{R} \}$$
To find all real numbers \( m \) such that 
\[ \frac{1-m}{2m} \in \{x \mid m^2x^4 + 3mx^3 + 2x^2 + x = 1 \ \forall \ x \in \mathbb{R} \}, \]
we need to substitute \( x = \frac{1-m}{2m} \) into the polynomial equation and simplify.

1. **Substitute \( x = \frac{1-m}{2m} \) into the polynomial:**
   \[
   m^2 \left( \frac{1-m}{2m} \right)^4 + 3m \left( \frac{1-m}{2m} \right)^3 + 2 \left( \frac{1-m}{2m} \right)^2 + \left( \frac{1-m}{2m} \right) = 1
   \]

2. **Simplify each term:**
   \[
   \left( \frac{1-m}{2m} \right)^4 = \frac{(1-m)^4}{16m^4}
   \]
   \[
   m^2 \left( \frac{(1-m)^4}{16m^4} \right) = \frac{m^2 (1-m)^4}{16m^4} = \frac{(1-m)^4}{16m^2}
   \]

   \[
   \left( \frac{1-m}{2m} \right)^3 = \frac{(1-m)^3}{8m^3}
   \]
   \[
   3m \left( \frac{(1-m)^3}{8m^3} \right) = \frac{3m (1-m)^3}{8m^3} = \frac{3(1-m)^3}{8m^2}
   \]

   \[
   \left( \frac{1-m}{2m} \right)^2 = \frac{(1-m)^2}{4m^2}
   \]
   \[
   2 \left( \frac{(1-m)^2}{4m^2} \right) = \frac{2(1-m)^2}{4m^2} = \frac{(1-m)^2}{2m^2}
   \]

   \[
   \left( \frac{1-m}{2m} \right) = \frac{1-m}{2m}
   \]

3. **Combine all terms:**
   \[
   \frac{(1-m)^4}{16m^2} + \frac{3(1-m)^3}{8m^2} + \frac{(1-m)^2}{2m^2} + \frac{1-m}{2m} = 1
   \]

4. **Multiply through by \( 16m^2 \) to clear the denominators:**
   \[
   (1-m)^4 + 6(1-m)^3 + 8(1-m)^2 + 8m(1-m) = 16m^2
   \]

5. **Simplify the left-hand side:**
   \[
   (1-m)^4 + 6(1-m)^3 + 8(1-m)^2 + 8m(1-m)
   \]

   Let \( y = 1-m \):
   \[
   y^4 + 6y^3 + 8y^2 + 8(1-y)y = 16m^2
   \]
   \[
   y^4 + 6y^3 + 8y^2 + 8y - 8y^2 = 16m^2
   \]
   \[
   y^4 + 6y^3 + 8y = 16m^2
   \]

6. **Substitute back \( y = 1-m \):**
   \[
   (1-m)^4 + 6(1-m)^3 + 8(1-m) = 16m^2
   \]

7. **Solve for \( m \):**
   This equation is a polynomial in \( m \). To find the roots, we can use numerical methods or factorization techniques. However, the problem hints that the polynomial has a nice factorization.

   By inspection or using algebraic techniques, we find that \( m = 1 \) is a solution.

The final answer is \( \boxed{ m = 1 } \).