 Solution

If $n$ is positive and odd, $n^2$ is odd. The function $x\mapsto x^{n^2}$ is a strictly increasing function on the real numbers, so it is injective. Every real $y$ has the real odd root $\mathrm{sgn}(y)|y{|}^{1/n^2}$, so the function is also surjective.

If $n$ is positive and even, the exponent is positive and even. Then $f(1)=f(-1)$, while $f(x)\ge 0$ for all real $x$. These facts disprove injectivity and surjectivity, respectively. ThereforeIf the convention $0\in \mathbb{N}$ is used, $n=0$ gives exponent zero. Interpreting $x^0$ as the constant polynomial $1$ makes the function neither injective nor surjective. If $0^0$ is left undefined, that case does not define the stated function on all the real numbers. Neither convention adds a qualifying value of $n$.