There is indeed an asymmetry (not anti symmetry) in the Doppler effect through a medium under exchange of the wave source and wave observer. What does stationary mean? With respect to whom? Those are questions that you will find yourself asking as you go deeper and deeper into physics.
This asymmetry comes from the fact that one usually considers either the emitter or the observer stationary with respect to the medium.
The equation for the (non-relativistic) Doppler effect reads:
$$ \nu = \nu_0 \frac{v \pm v_r}{v \mp v_s} .$$ with $v$ the speed of sound in the medium, $v_r$ the speed of the receiver and $v_s$ the speed of the source.
Let us say for the sake of simplicity, $v = 1$, and that the source is getting closer to an receiver with speed $v_s = 0.5$, so $$ \nu = \nu_0 \frac{1}{1 - 0.5} = 2\nu_0 .$$
If we swap receiver and source $v_s \to -v_r$, $v_r \to -v_s$, one would expect the Doppler effect to remain unchanged. It is however, not $$\nu = \nu_0 \frac{1+0.5}{1} = \frac{3}{2} \nu_0$$
The problem here came from the fact that the speed of sound through the medium changed when changing the frame of reference. When swapping the source and receiver, we changed who is stationary with respect to the medium.
This is all an artifact consequence of Galilean relativity. After the Michelson-Morley experiment, Einstein arrived to a weird conclusion: The speed of light must be the same to any observer. This poses a change in perspective in how we see physics. Galilean relativity assumes (as anyone would, really) that after a change in the frame of reference space and time are left the same, and therefore the speed of propagation through the medium is allowed to change.
If you force the speed through 'the medium' to stay the same (this is to paint a picture, there is no 'medium'. Light does not need a medium to propagate), you must allow space and time to change. The way these quantities change is given by Lorentz transformations$$
t' = \gamma(t - vx) \\
x' = \gamma(x - vt),$$
with $\gamma = \frac{1}{1 - v^2/c^2}$ the Lorentz factor. The way this reads is: You are an observer in your own reference frame. You observe a certain event happening at some time t, at some position x. Another observer, who moves in your reference frame at speed $v$, will observe that same event to have happened at some time $t'$, at some position $x'$.
Ok. Going the Doppler effect now. The Doppler effect talks about waves and the easiest wave we can think about is the harmonic wave. A harmonic wave propagating at $c$, with angular frequency $\omega = 2\pi \nu$ and wave number $k = \frac{2\pi}{\lambda} = \frac{\omega}{c}$ is of the form
$$\psi(t, x) = sin(\omega t - kx + \phi_0).$$
If we want this wave to change correctly under change of frame of reference, we need to impose something like
$$\omega t - k x + \phi_0 = \omega' t' - k' x' + \phi_0'.$$
Since we know the relation between space and time for the different frames of reference from the Lorentz transformation, we can use that in our equation
$$\omega t - k x + \phi_0 = \omega' \gamma(t - vx) - k' \gamma(x - vt) + \phi_0'\\
= \gamma(\omega' + k'v) t - \gamma(\omega' v + k')x.$$
By comparison, we need (recalling that $k = \frac{\omega}{c}$) $$\omega = \gamma(\omega' + k'v) = \gamma \omega' (1 + v/c) $$
Solving for $\omega'$
$$\omega' = \frac{\omega}{\gamma (1 + \frac{v}{c})} = \omega\frac{\sqrt{1 - \frac{v^2}{c^2}}}{1 + v/c} = \omega \frac{\sqrt{(1-\frac{v}{c})(1 + \frac{v}{c})}}{\sqrt{(1-\frac{v}{c})^2}}=\omega \sqrt{\frac{1 + \frac{v}{c}}{1 - \frac{v}{c}}} = \omega \sqrt{\frac{c + v}{c - v}}$$
Or equivalently, $$\nu = \nu_0 \sqrt{\frac{c + v}{c - v}}.$$
This is your symmetric Doppler effect. Or better put, the relativistic Doppler effect.
By the way, if you are interested, under the assumption of what should and should not change under frame of reference, one can talk about many types of Doppler shifts. The most famous are (the just calculated) relativistic Doppler effect, the gravitational redshift, and cosmological redshift.
These are all consequences of something called 'covariance', or 'the fact that physics is unimpressed by your choice of coordinates'.
