Consider a semi direct product of two finite groups $G=N\rtimes H$, where $|N|$ and $|H|$ are coprime, and denote $S$ the set of elements in $G$, whose order divides $|H|$.
What can we say about these elements? As an easy example we can see that if $N$ and $H$ commute and we have a direct product, then $S=H$. Similarly (or as a consequence of the previous observation) we can see that if the semidirect product is non-trivial, then a non-trivial element $(n,h) \in S$ cannot satisfy $n \in Z(N \rtimes H)$, as $(n,h)^k = (n^k, h^k)$ and $n^k$ is not the identity.
In the case of a semi-direct product $C_q \rtimes C_p$ of cyclic groups of prime orders $p,q$ however, I think $S$ is exactly $G - C_q$. As an example of the reasoning behind this claim, consider the element $(1,1)$ (I am using additive notation).
$(1,1)^p = (1,1)(1,1)^{p-1}=(1 + \varphi(1),2)(1,1)^{p-2}= ... = (1 + \varphi(1) + \varphi^2(1) + ... + \varphi^{p-1}(1), 0)$, where $\varphi$ is a non-trivial automorphism of order $p$.
Such an automorphism can be described by $1 \mapsto a$, where $a^p =1$ (because $\varphi^p = id$, and so $1=\varphi^p(1) = a^p$), and so we have $(1 + \varphi(1) + \varphi^2(1) + ... + \varphi^{p-1}(1) ) = 1 + a + a^2 + ... +a^{p-1}$. This is a geometric sum, and thus it can be expressed as $(a^p -1) / (a-1)$, which is $0$ in $\mathbb{Z_q}$, and so $(1,1)^p = (0,0)$. A similar argument can be applied on other elements in $G - C_q$.
I have not spent much time on this question, but I would appreciate any input concerning this question. Some other things I have wondered about - if the semi-direct product is nontrivial, is $S$ always strictly larger than $H$? If $S$ is strictly larger than $H$, does this give us any information on the amount of groups $\tilde{H} \neq H$ that complement $N$ (i.e. $N \rtimes H = N \rtimes \tilde{H}$, but $H \neq \tilde{H}$) ? Note - if the semi-direct product is non-trivial, there must always exist a different complement - this is because $H^g$ is still a complement, and if all $H^g = H$, then $H$ is normal in $G$ and so the semi-direct product is a direct product.
Every element of $S$ is contained in a conjugate of $H$ (by the Schur-Zassenhaus Theorem), so $S$ is the union of the set of conjugates of $H$ in $G$.
Yes, the semidirect product is nontrivial if and only if $S$ strictly contains $H$, but you seem to have answered that yourself.