Integrals revisited in 2019.

Performing the substitution ${ x }^{ \frac { 1 }{ k } } = t$ , the given conditions become

$\large\ \displaystyle \int _{ 0 }^{ 1 }{ { \left( f\left( t \right) \right) }^{ n - k }{ t }^{ k - 1 }dt } = \frac { 1 }{ n }$ , $k=1, 2, ..., n-1.$

Observe that this equality also holds for $k = n$ . With this in mind we write

$\large\ \int _{ 0 }^{ 1 }{ { \left( f\left( t \right) -t \right) }^{ n-1 }dt } = \displaystyle \sum _{ k=1 }^{ n }{ \displaystyle { \left( -1 \right) }^{ k-1 }\begin{pmatrix} n-1 \\ k-1 \end{pmatrix} } \int _{ 0 }^{ 1 }{ { \left( f\left( t \right) \right) }^{ n-k }{ t }^{ k-1 }dt } = \displaystyle \sum _{ k=1 }^{ n }{ { \left( -1 \right) }^{ k-1 }\begin{pmatrix} n-1 \\ k-1 \end{pmatrix} } \frac { 1 }{ n } = \frac { 1 }{ n } { \left( 1-1 \right) }^{ n-1 } = 0$

Because $n-1$ is even, ${ \left( f\left( t \right) - t \right) }^{ n-1 } \ge 0$ . The integral of this function can be zero only if $f (t) - t = 0$ for all $t \in [0, 1].$ Hence the only solution to the problem is $f(x) = x.$