Fourier Transform Time Scaling
Let $y(t) = x(t^2)$, and $a$ be such that:

$$
\begin{align}
  y(at) 
   &amp;= x((at)^2) \\
   &amp;= x(a^2t^2) \\
   &amp;= x(bt^2)
\end{align}
$$
In other words, $a^2 = b$.

The Fourier Transform time-scaling property states that given the Fourier Transform pair:
$$
\begin{align}
  y(t) \leftrightarrow Y(w)
\end{align}
$$
and $a&gt;0$, we have the relation:
$$
\begin{align}
  y(at) &amp;\leftrightarrow \frac{1}{a}Y(\frac{w}{a})
\end{align}
$$

Thus,
$$
\begin{align}
  x(bt^2) &amp;\leftrightarrow \frac{1}{a}Y(\frac{w}{a}) \\
          &amp;\leftrightarrow \frac{1}{\sqrt{b}}Y(\frac{w}{\sqrt{b}})
\end{align}
$$

Now, if we define
$$
\begin{align}
Y(w) = G(w^2)
\end{align}
$$
Then,
$$
\begin{align}
Y(w/a) &amp;= G(w^2/a^2) \\
Y(\frac{w}{\sqrt{b}}) &amp;= G(\frac{w^2}{b})
\end{align}
$$

Yielding,
$$
\begin{align}
x(bt^2) &amp;\leftrightarrow \frac{1}{a}Y(\frac{w}{a}) \\
        &amp;\leftrightarrow \frac{1}{\sqrt{b}}G(\frac{w^2}{b}) \\
        &amp;\leftrightarrow \frac{1}{\sqrt{b}}G((\frac{w}{\sqrt{b}})^2)
\end{align}
$$
Or, in prose (1/sqrt(b)) times a function of (w/sqrt(b))^2.