 Solution

If $\gcd (m,|H|)=1$, choose $r$ with $mr\equiv 1(\mathrm{mod}|H|)$ by the Bezout identity. The Lagrange theorem gives $x^{|H|}=1$ for every $x\in H$, so $(x^m{)}^r=x$ and $(x^r{)}^m=x$. Thus $x\mapsto x^m$ is bijective, even though it need not be a homomorphism.

Conversely, if a prime $p$ divides both $m$ and $|H|$, the Cauchy theorem for groups gives $x\ne 1$ with $x^p=1$. Then $x^m=1$, so the power map sends both $x$ and the identity to the identity and is not injective. This proves the power-map criterion for a finite group.