I'm not able to understand what conjugate of a transform means. In my textbook it says that the fourier inverse transform is just the conjugate, $\mathcal{F}^*\mathcal{F} = I$. Though I'm not sure how I can use this...
suppose I'm given $\mathcal{F}[x(t)] = X(\omega)$, how can we calculate $\mathcal{F}^{-1}[x(\omega)]$ using this?
In operatorial sense we have the Fourier integral operator $$ (\mathcal Ff)(x)=\int_{-\infty}^\infty f(y)\,k(x,y) \,\mathrm dy=\hat f(x) $$ where the kernel, according to the sign convention of your book, is $$k(x,y)=\frac{\mathrm e^{ixy}}{\sqrt{2\pi}}$$ The inverse Fourier integral operator is then $$ (\mathcal F^{-1}\hat f)(x)= f(x)=\int_{-\infty}^\infty \hat f(y)\,k^*(x,y) \,\mathrm dy=(\mathcal F^*\hat f)(x)=(\mathcal F^*\mathcal F f)(x) $$ and then $\mathcal F^{-1}=\mathcal F^{*}$ and $\mathcal F^{*}\mathcal F=\mathcal I$.
Actually this is not useful to find the inverse (pay attention on the variables). This is also a consequence of the duality property of Fourier transform usually written $$ X(\omega)=\mathcal F\{x(t)\}\quad\Longrightarrow \quad\mathcal F\{X(t)\}=x(-\omega) $$ so that \begin{align} \mathcal F^*\{X(t)\}=\mathcal F^*\{\mathcal F\{x(\omega)\}\}&=\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}X(t)\mathrm e^{i\omega t}\,\mathrm d t\right)^*\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}X^*(-t)\mathrm e^{i\omega t}\,\mathrm d t\\ &=\mathcal F\{X^*(-t)\}=x(\omega) \end{align} that is $\mathcal F^*\{\mathcal F\{x(\omega)\}\}=x(\omega)$, or $\mathcal F^{*}\mathcal F=\mathcal I$.