In the Wikipedia article on the twin pararadox, there is an interesting chapter which calculates the difference of age for the twins, with steps of accelerated movement, and steps with constant speed, for the traveller twin.
We simplify here the Wikipedia example in deleting the constant speed steps (so $T_c=0$), we have :
$$ \Delta \tau = \frac{4c}{a} {\rm arsinh} (\frac{aT_a}{c}) \qquad, \Delta t = 4 T_a\tag{1}$$ where $a$ is the proper acceleration of the traveller twin, relatively to the sedentary twin, $T_a$ is the (sedentary twin) time necessary for the traveller twin to go from a speed zero to a maximum speed (and vice-versa), $ \Delta \tau$ is the elapsed time for the traveller twin, $ \Delta t$ is the elapsed time for the sedentary twin. We have : $\Delta \tau < \Delta t$
Now, certainly, there must be an asymetry about proper acceleration, that is : the proper acceleration of the sedentary twin, seen by the traveller twin, has to be different (in absolute value) with $a$.
I am looking for a rigourous argument, and I have only some hints. I think there is a symmetry between relative speeds $\vec v_{S/T} = -\vec v_{T/S}$, and this should extend to the spatial part of the $4$-velocities $\vec u_{S/T} = -\vec u_{T/S}$, while the time part of the $4$-velocities $u_{S/T}^0 =u_{T/S}^0$should be equal. The problem, I think, seems to come from the definition of the proper acceleration $ \vec a = \frac {d \vec u}{dT}$, where $T$ is the time of the observer. The symmetry would be broken, because, in one case (proper acceleration of the traveller twin), one hase to use $dT=dt$, while in the other case (proper acceleration of the sedentary twin), one has to use $dT=d \tau$, so, there would not exist a symmetry between $\vec a_{T/S} = \frac {d \vec u_{T/S}}{dt}$ and $\vec a_{S/T} = \frac {d \vec u_{S/T}}{d\tau}$
Is this correct ?
