Einstein’s special relativity theory postulates that, as long as we are dealing with uniform motion, there is no frame of reference preferable over another. Also, all the laws of physics (and thus electromagnetism) must be valid irrespective of the chosen reference frame. One immediate conclusion then must be that we need not differentiate between a moving source (with stationary receiver) and a moving receiver (with stationary source) as in the classical case with a medium.
Again, we start with the plane wave, represented by its phasor $A(t)$, as in the classical case.
$$A(\mathbf{r},t)=A_{0}\exp\bigl(i\omega t-i\mathbf{k}\cdot\mathbf{r}\bigr) \label{plane-wave}$$
where $A_0$ describes the (constant) amplitude and phase of the emitted signal, $\omega$ is its frequency at the source and $\mathbf{k}$ is the wave vector, describing the spatial frequency of the wave and its direction.
In the reference frame of the source, a moving receiver can be described similarly to the classical case, with the wave field at the location of the receiver moving with velocity $\mathbf{v}$ (in the source frame of reference) given by
$$A(\mathbf{r},t)=A_{0}\exp\Bigl[i\omega t-i\mathbf{k}\cdot\bigl(\mathbf{r}_{0}+\mathbf{v} t\bigr)\Bigr] \label{field-R}$$
In the classical case, the received frequency was determined by differentiation of the phase of the field with respect to time,
$$\begin{aligned}
\omega^\prime_{R} &= \partial_t \arg\bigl[A(\mathbf{r},t)\bigr] \vphantom{\biggl(\biggr)}\\
& =\omega-\mathbf{k}\cdot\mathbf{v}\\
& =\omega\biggl(1-\frac{\mathbf{\hat{k}}\cdot\mathbf{v}}{c}\biggr)
\end{aligned}\label{phase-differentiation}$$
In relativistic physics, however, in order to obtain the frequency observed by the receiver we have to do the phase differentiation in the reference frame of the receiver. Einstein postulated that we need to apply the Lorentz transformation to convert the measurements of time and space in a reference frame which is defined to be at rest (source) into the measurements of the same quantities that would be obtained in another reference frame (receiver) which is moving with velocity $\mathbf{v}$ relative to the frame at rest. The Lorentz transforms for time and space are
$$\begin{gathered}
t^\prime = \gamma \biggl(t - \frac{\mathbf{v}\cdot \mathbf{r}}{c^2} \biggr)\\
\mathbf{r}^\prime = \mathbf{r} + \mathbf{v} \bigl(\gamma - 1 \bigr) \frac{\mathbf{\hat{v}} \cdot \mathbf{r}}{\mathbf{v} \cdot \mathbf{v}} - \gamma \, \mathbf{v} \, t
\end{gathered}\label{Lorentz-transformation}$$
where the primed variables denote the quantities in the moving reference frame and
$$\gamma = \biggl(1 - \frac{\mathbf{v}\cdot\mathbf{v}}{c^2}\biggr)^{-\frac{1}{2}}$$
is the Lorentz factor. The inverse transforms are
$$\begin{gathered}
t = \gamma \biggl(t^\prime + \frac{\mathbf{v}\cdot \mathbf{r}^\prime}{c^2} \biggr)\\
\mathbf{r} = \mathbf{r}^\prime + \mathbf{v} \bigl(\gamma - 1 \bigr) \frac{\mathbf{\hat{v}} \cdot \mathbf{r}^\prime}{\mathbf{v} \cdot \mathbf{v}} + \gamma \, \mathbf{v} \, t^\prime
\end{gathered}\label{inverse-Lorentz-transformation}$$
The difference in the forward and inverse transforms is merely in the sign of $\mathbf{v}$ in agreement with the postulated equivalence of all reference frames. It is equally valid to define the receiver reference frame to be at rest with the source frame moving in the opposite direction $-\mathbf{v}$ (assuming the coordinate systems in both cases to have the same orientation). Hence the inverse transform can only differ in the sign of the velocity vector.