Assume $G$ is isomorphic to $G'$ and $H$ is a subgroup of $G$ with $|H|=d$.
Show that there exists $H'\leq G'$ s.t. $|H'|=d$.
Let $f: G \longrightarrow G'$ be an isomorphism. Then, there exists $H' \subseteq G'$ defined as $$ H':=\left\{ f(h), \space \forall h \in H \right\} $$ and since $f$ is injective, $|H'|=d$. To show that $H'$ is a subgroup of $G'$:
$\bullet \space$ Closure: $\forall h_1,h_2 \in H:$ $$ h_1 \cdot h_2 \in H \iff f(h_1 \cdot h_2) \in f(H) \iff f(h_1) \cdot f(h_2) \in f(H) \iff h_1'\cdot h_2' \in H' \quad\checkmark $$
$\bullet \space$ Identity: $$ e \in H \iff f(e) \in f(H) \iff e' \in H' \quad \checkmark $$
$\bullet \space$ Inverses: $$ h^{-1} \in H \iff f(h^{-1}) \in f(H) \iff f^{-1}(h) \in f(H) \iff h'^{-1}\in H' \quad \checkmark $$
Therefore, there exists $H' \leq G'$ with $|H'|=d$.
Let $f\colon G\rightarrow G'$ be a group isomorphism. By assumption such an $f$ exists. Then define $H':=f(H)$. Because $f$ is a homomorphism, $f(H)$ is a subgroup of $G'$, hence a group. Since the kernel of $f=f_{\mid H}$ is trivial, because $f$ is injective, we have $$ H\cong H/\ker(f)\cong f(H). $$ Hence $|H|=|H'|=d$.