A family of examples is given by Keith Kearnes in a comment, which I will put here as an answer.
Let $V$ be a finite-dimensional $\mathbf F_p$-vector space with dimension $d \geq 1$. Set $G = V \rtimes {\rm Aut}(V)$, with group law
$$
(v,A)(w,B) = (v + A(w),AB)
$$
and $(v,A)^{-1} = (-A^{-1}(v),A^{-1})$.
Theorem. With notation as above, let $L$ be a $1$-dimensional subspace of $V$, $I$ be the identity in ${\rm Aut}(V)$, and $H = (L,I)$. Then $H$ is a subgroup of $G$ and
$$
\bigcup_{g \in G} gHg^{-1} = (V,I) = \{(v,I) : v \in V\},
$$
which is a subgroup of $G$ and is strictly larger than $H$ when $d > 1$.
Proof. Let $v_0$ be a nonzero vector in $L$, so $L = \mathbf F_pv_0$ and $H = \langle (v_0,I)\rangle$. For $g = (v,A) \in G$,
$$
g(v_0,I)g^{-1} = (A(v_0),I),
$$
where the $v$ in $g$ has no effect in the end.
Thus $gHg^{-1} = g\langle(v_0,I)\rangle{g}^{-1} = \langle (A(v_0),I)\rangle \subset (V,I)$. Since ${\rm Aut}(V) = {\rm GL}(V)$, $\bigcup_{g \in G} g(v_0,I)g^{-1} = \left(\bigcup_{A \in {\rm GL}(V)} A(v_0),I\right) = (V - \{0\}, I)$ and $(0,I)$ is the identity in $H$, so
$$
\bigcup_{g \in G} gHg^{-1} = (V,I),
$$
which is a subgroup of $G$. It is strictly larger than $H$ unless $L = V$, which means $V$ is $1$-dimensional. So as long as $V$ has dimension greater than $1$, $\bigcup_{g \in G} gHg^{-1} \not= H$. QED
In the proof, we relied on the property that ${\rm Aut}(V)$ acts transitively on $V - \{0\}$. Every nontrivial finite group whose automorphisms act transitively on the non-identity elements in it is a finite-dimensional vector space over $\mathbf F_p$ for some prime $p$: see Theorem $3.15$ here.
We can view the abstract semi-direct product $V \rtimes {\rm Aut}(V)$ more concretely as the set of affine-linear maps $V \to V$ under composition, where $(v,A)$ in the semi-direct product corresponds to the affine-linear map $\mathbf x \mapsto A\mathbf x + v$ on $V$.
The smallest example of such $V$ is $\mathbf F_2^2$, for which $V \rtimes {\rm Aut}(V)$ has size $4 \cdot 6 = 24$. Each affine-linear map $\mathbf F_2^2 \to \mathbf F_2^2$ is a permutation of $\mathbf F_2^2$, a set of size $4$, so the affine-linear maps being a group of order $24$ implies this group is isomorphic to $S_4$. Thus $S_4$ is an example of the kind of group $G$ being sought, with $\mathbf F_2^2$ embedded in $S_4$ as the translations $\{\mathbf x \mapsto \mathbf x + v : v \in \mathbf F_2^2\}$ on $\mathbf F_2^2$, where the nonzero translations are the three permutations in $S_4$ of type $(2,2)$. That leads to the following concrete example of $G$ and $H$.
Example. Let $G = S_4$ and $H = \langle (12)(34)\rangle$, which has order $2$. Then
$$
\bigcup_{g \in G} gHg^{-1} = \{(1),(12)(34),(13)(24),(14)(23)\}
$$
is the normal subgroup of $S_4$ with order $4$.
When $V$ has dimension greater than $1$ and $B$ is a subgroup of ${\rm Aut}(V)$ that acts transitively on $V - \{0\}$, the proof of the theorem above works for the group $G = V \rtimes B$: its subgroup $H = (L,I)$ has the property that $\bigcup_{g \in G} gHg^{-1} = (V,I)$, which is a subgroup of $G$ strictly larger than $H$.
Example.
Taking $V = {\mathbf F}_2^2$ again, the group ${\rm Aut}(V) \cong {\rm GL}_2(\mathbf F_2) \cong S_3$ has a subgroup $B$ of order $3$ that acts transitively on the nonzero elements of $V$: $B = \{(\begin{smallmatrix}1&0\\0&1\end{smallmatrix}), (\begin{smallmatrix}1&1\\1&0\end{smallmatrix}), (\begin{smallmatrix}0&1\\1&1\end{smallmatrix})\}$. Therefore in $V \rtimes B$ there is a subgroup $H$ of order $2$ such that $\bigcup_{g \in G} gHg^{-1} = (V,I)$, which is a subgroup of $G$ larger than $H$. The group $G$ is isomorphic to $A_4$, with $H$ corresponding to a subgroup of $A_4$ generated by a permutation of type $(2,2)$ and $\bigcup_{g \in G} gHg^{-1}$ corresponding to the unique subgroup of $A_4$ with order $4$.
