= 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 $\operatorname{sgn}(y)|y|^{1/n^2}$, so the <function> is also <surjective>.
If $n$ is positive and even, the exponent is positive and even. Then $f(1)=f(-1)$, while $f(x)\geq0$ for all real $x$. These facts disprove <injectivity> and <surjectivity>, respectively. Therefore
$$
\boxed{f\text{ is onto iff }n\text{ is odd};\qquad f\text{ is one-to-one iff }n\text{ is odd}.}
$$
If 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 $n$.
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