Fourier Transforms, symmetry, real/imaginary
Your suspicion is correct. You can show this as follows:

$$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\tag{1}$$

$$X^*(\omega)=\int_{-\infty}^{\infty}x^*(t)e^{j\omega t}dt\tag{2}$$

Now if $x(t)$ is purely imaginary we have $x^*(t)=-x(t)$, and, consequently,

$$X^*(\omega)=-\int_{-\infty}^{\infty}x(t)e^{j\omega t}dt=-X(-\omega)\tag{3}$$