Find all real $a,b,c,d,e,f$ that satisfy the system
$4a = (b + c + d + e)^4$
$4b = (c + d + e + f)^4$
$4c = (d + e + f + a)^4$
$4d = (e + f + a + b)^4$
$4e = (f + a + b + c)^4$
$4f = (a + b + c + d)^4$
1. Denote \( a, b, c, d, e, f \) by \( a_0, a_1, a_2, a_3, a_4, a_5 \) respectively, with indices taken modulo 6. This means \( a_6 = a_0 \), \( a_7 = a_1 \), and so on.
2. Given the system of equations:
   \[
   4a_0 = (a_1 + a_2 + a_3 + a_4)^4
   \]
   \[
   4a_1 = (a_2 + a_3 + a_4 + a_5)^4
   \]
   \[
   4a_2 = (a_3 + a_4 + a_5 + a_0)^4
   \]
   \[
   4a_3 = (a_4 + a_5 + a_0 + a_1)^4
   \]
   \[
   4a_4 = (a_5 + a_0 + a_1 + a_2)^4
   \]
   \[
   4a_5 = (a_0 + a_1 + a_2 + a_3)^4
   \]
3. Let \( S = a_0 + a_1 + a_2 + a_3 + a_4 + a_5 \). We can rewrite each equation in terms of \( S \):
   \[
   4a_i = (S - a_i - a_{i+1})^4
   \]
4. Assume \( a_i \) are positive. We can express \( S \) in terms of \( a_i \):
   \[
   S = \sqrt[4]{4a_i} + a_i + a_{i-1}
   \]
5. Subtracting the equations for \( i \) and \( i-1 \):
   \[
   \sqrt[4]{4a_i} + a_i + a_{i-1} - (\sqrt[4]{4a_{i-1}} + a_{i-1} + a_{i-2}) = 0
   \]
   Simplifying, we get:
   \[
   \sqrt[4]{4a_i} - \sqrt[4]{4a_{i-1}} = a_{i-2} - a_i
   \]
6. If \( a_i < a_{i-1} \), then \( a_{i-2} < a_i \). This implies \( a_{i-1} > a_{i-2} \). Continuing this logic, we see that \( a_{i-k} \) is alternatively greater than the greatest or smaller than the smallest of the numbers \( a_i, a_{i-1}, \dots, a_{i-(k-1)} \). Since we eventually come back to \( a_i \), this is a contradiction.
7. Therefore, all the numbers must be equal. Let \( a_i = k \) for all \( i \). Substituting into the original equations:
   \[
   4k = (4k)^4
   \]
   \[
   4k = 256k^4
   \]
   \[
   1 = 64k^3
   \]
   \[
   k^3 = \frac{1}{64}
   \]
   \[
   k = \frac{1}{4}
   \]
8. Thus, \( a = b = c = d = e = f = \frac{1}{4} \).

The final answer is \( \boxed{\frac{1}{4}} \).