Functions with $ f \bigg( p \Big( f \big( p ( x ) \big) \Big) \bigg) = p \bigg( f \Big( p \big( f ( x ) \big) \Big) \bigg) $ for all polynomials $ p $ 
Find all functions $ f : \mathbb R \to \mathbb R $ such that
$$ f \bigg( p \Big( f \big( p ( x ) \big) \Big) \bigg) = p \bigg( f \Big( p \big( f ( x ) \big) \Big) \bigg) $$
for all $ x \in \mathbb R $ and all nonconstant polynomials $ p : \mathbb R \to \mathbb R $.

Obviously, the identity function is a solution.
If constant $ p $ was allowed, identity would be the only solution, as for any $ x \in \mathbb R $, we could take $ p $ to be the constant function with the value $ x $, and the functional equation would give $ f ( x ) = x $.
Source:
Number five on the list at the end of this page.
 A: Not only it can be shown that identity is the only solution, but also one can only require the functional equation to hold for nonconstant affine functions, and still get the same answer.
For any $ y , z \in \mathbb R $ with $ z \ne 0 $, the function $ p : \mathbb R \to \mathbb R $ defined with $ p ( x ) = z x + y $ for all $ x \in \mathbb R $ is a nonconstant polynomial. Considering only the functions of this form, the functional equation reads
$$ f \big( z f ( z x + y ) + y \big) = z f \big( z f ( x ) + y \big) + y $$
for all $ x , y , z \in \mathbb R $ with $ z \ne 0 $.
Assuming $ f ( 0 ) \ne 0 $, we can substitute $ 0 $ for both $ x $ and $ y $ and $ \frac z { f ( 0 ) } $ for $ z $ to get $ f ( 0 ) f ( z ) = z f ( z ) $, which shows that for all $ z \in \mathbb R \setminus \{ 0 , f ( 0 ) \} $ we have $ f ( z ) = 0 $. But setting $ y = 0 $ and $ x = z = | f ( 0 ) | + 1 $, we will have $ x , z x + y \in \mathbb R \setminus \{ 0 , f ( 0 ) \} $, which gives $ f ( x ) = f ( z x + y ) = 0 $, and thus $ f ( 0 ) = \big( | f ( 0 ) | + 1 \big) f ( 0 ) $, which contradicts $ f ( 0 ) \ne 0 $. Therefore, we must have $ f ( 0 ) = 0 $.
Setting $ x = 0 $ and $ y = z = 1 $ we get $ f \big( f ( 1 ) + 1 \big) = f ( 1 ) + 1 $, which show that there is some $ a \in \mathbb R $ (equal to either $ 1 $ or $ f ( 1 ) + 1 $) such that $ f ( a ) \ne 0 $. Hence, substituting $ 0 $ for $ x $, $ a $ for $ y $ and $ \frac { z - a } { f ( a ) } $ for $ z $, we get $ f ( z ) = z $ for all $ z \in \mathbb R \setminus \{ a \} $. Now, we can choose some $ b \in \mathbb R \setminus \{ a \} $ with $ f ( b ) \ne 0 $, repeat the argument to get $ f ( z ) = z $ for all $ z \in \mathbb R \setminus \{ b \} $, and thus conclude that $ f $ must be the identity function.
