general relativity - S-duality of Einstein-Maxwell-Dilaton theory

Consider theory with action

$$S = \int d^D x \sqrt{-g} (R - \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2k!} e^{a \phi} F^2 _{[k]} )$$

where $\phi$ is dilaton and $F_{[k]}$ is electromagnetic $k$-form.

S-duality is the symmetry of this action

$$g_{\mu \nu} \to g_{\mu \nu} \ , \ \ F \to e^{- a \phi} \star F \ , \ \ \phi \to - \phi$$

I cannot understand why we have to use this transformation in order to get , for example, magnetic solution if electric one is already known. Why cannot we use only $F \to \star F \ , \ \ \phi \to \phi$ transformation?

Moreover, the equations of motion for magnetic solution are

$$\partial _\mu (\sqrt{-g} e^{a \phi} F^ {\mu \alpha_2 ... \alpha_k } ) = 0$$ And it is claimed that magnetic solution of this equation (for diagonal radially symmetric metric) is

$$F_{[k]} = \frac{P}{R^{D-2}} d\theta_1 \wedge ... \wedge d\theta_k$$

But I cannot understand why it does not depend on dilaton via $e^{- a \phi}$ as electric solution does.