Effects of interchanging sine terms with cosine terms
If you have a $T$-periodic signal $x(t)$ with a Fourier series expansion

$$x(t)=a_0+\sum_{n=1}^{\infty}\left[a_n\cos(n\omega_0t)+b_n\sin(n\omega_0t)\right],\quad \omega_0=\frac{2\pi}{T}\tag{1}$$

then its Hilbert transform is given by

$$\mathcal{H}\{x(t)\}=\hat{x}(t)=\sum_{n=1}^{\infty}\left[a_n\sin(n\omega_0t)-b_n\cos(n\omega_0t)\right]\tag{2}$$

because $\mathcal{H}\{\cos(t)\}=\sin(t)$, $\mathcal{H}\{\sin(t)\}=-\cos(t)$, and $\mathcal{H}\{c\}=0$ for constant $c$.

This is not exactly the same as just swapping the coefficients $a_n$ and $b_n$. The signal with swapped coefficients (assuming $a_0=0$) is

$$y(t)=\sum_{n=1}^{\infty}\left[a_n\sin(n\omega_0t)+b_n\cos(n\omega_0t)\right]$$

which is related to $\hat{x}(t)$ by

$$y(t)=-\hat{x}(-t)\tag{3}$$

It is also equal to the Hilbert transform of $x(-t)$:

$$y(t)=\mathcal{H}\{x(-t)\}\tag{4}$$

So swapping the coefficients is equivalent to applying a linear time-varying system to the signal. It cannot be achieved by a Hilbert transform alone (which is a time-invariant operation).

Note that for even signals ($b_n=0$) and for odd signals ($a_n=0$) one does indeed obtain the Hilbert transform (for odd signals with a negative sign) by swapping coefficients. This is shown in the example in hops's answer.