= Solution
Let $L$ be the <subgroup> of $H\operatorname{wr}A_m$ generated by all its elements of order $3$. The top <group> $A_m$ is generated by 3-cycles, also when $m=3$, so $A_m\le L$.
Take $\sigma=(1\,2\,3)$. For any $h\in H$, the conjugate $h_1\sigma h_1^{-1}$ has order $3$, as does $\sigma^{-1}$. Thus their product belongs to $L$ and equals
$$
d_{12}(h)=h_1\sigma h_1^{-1}\sigma^{-1}=h_1h_2^{-1}.
$$
Using $\sigma^{-1}$ instead supplies $d_{13}(h)=h_1h_3^{-1}$. Since distinct coordinate groups commute,
$$
[d_{12}(h),d_{13}(k)]=[h,k]_1\in L.
$$
The hypothesis that $H$ is perfect says that products of its commutators generate $H$, so $H_1\le L$. Transitivity of $A_m$ now gives $H_i\le L$ for every coordinate. Hence $H^m\le L$ and $A_m\le L$, proving
$$
\boxed{H\operatorname{wr}A_m\text{ is generated by elements of order }3.}
$$
The proof does not require $H$ itself to be finitely generated, nor does it require $A_m$ to be perfect for $m=3,4$.
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