$\delta$ would scale inversely proportionally to the scaling of $y_t$.
Take the original model
\begin{aligned}
y_t &= \beta y_{t-1} + \delta h_t + \epsilon_t, \\
h_t &= a_0 + a_1 \epsilon^2_{t-1} + b_1 h_{t-1}, \\
\epsilon_t &\sim i.i.N(0,h_t) \\
\end{aligned}
and scale $y_t$ with a positive constant $c$. This turns the model into
\begin{aligned}
(cy_t) &= \beta (cy_{t-1}) + \frac{\delta}{c} (c^2 h_t) + (c\epsilon_t), \\
(c^2 h_t) &= (c^2 a_0) + a_1 (c\epsilon_{t-1})^2 + b_1 (c^2 h_{t-1}), \\
(c\epsilon_t) &\sim i.i.N(0,c^2 h_t) \\
\end{aligned}
which can be expressed as
\begin{aligned}
\tilde y_t &= \beta \tilde y_{t-1} + \tilde\delta \tilde h_t + \tilde\epsilon_t, \\
\tilde h_t &= \tilde a_0 + a_1 \tilde\epsilon_{t-1}^2 + b_1 \tilde h_{t-1}, \\
\tilde\epsilon_t &\sim i.i.N(0,\tilde h_t) \\
\end{aligned}
where
\begin{aligned}
\tilde y_t &= c y_t, \\
\tilde\delta &= \frac{\delta}{c}, \\
\tilde\epsilon_t &= c\epsilon_t, \\
\tilde a_0 &= c^2 a_0 \quad \text{and} \\
\tilde h_t &= c^2 h_t.
\end{aligned}
The second line of the equation block above is what you are interested in.