= Solution
The given <unitriangular group of degree three over F3> has $27$ elements, just as $G$ does, so counting alone cannot distinguish them. Instead compare <element orders>.
Write an element as $I+U$, with $U$ strictly upper triangular of size $3$. Then $U^3=0$. Over the <finite field> $\mathbb F_3$, the <binomial theorem for commuting matrices> gives
$$
(I+U)^3=I+3U+3U^2+U^3=I.
$$
Every nonidentity element of $H$ consequently has <element order> $3$. In contrast, $x\in G$ has <element order> $9$, by part (b). An <isomorphism> preserves <element order>, so
$$
\boxed{H\not\cong G.}
$$
Equivalently, the <exponent of a finite group> is $3$ for $H$ and $9$ for $G$.
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