The flat FLRW metric written in terms of conformal time $\eta$ is given by $$ds^2=a^2(\eta)(d\eta^2-dx^2-dy^2-dz^2)\tag{1}$$ where the scale factor $a(0)=1$. The interval of proper time $\tau$ measured by a comoving observer is given by $$d\tau=a(\eta)d\eta.$$ The coordinate system used to describe the metric in Eqn.(1) is that of a stationary observer fixed at the origin $(\eta,x,y,z)=(0,0,0,0)$ with four-velocity $$U^\mu=(1,0,0,0).$$ Assume that he observes a particle moving along a nearby path $x^\mu(\lambda)$ with four-velocity $$V^\mu=\frac{dx^\mu}{d\lambda}.$$ The four-momentum of the particle is $$P^\mu=mV^\mu.$$ The energy $\epsilon$ of the particle as measured by the stationary observer fixed at the origin $\eta=0$ is given by $$\epsilon=U_\mu P^\mu.$$ Now let us assume that the path $x^\mu(\lambda)$ is a geodesic. Using the metric in Eqn.(1) it can be shown that the covariant derivative of the energy $\epsilon$ along $x^\mu(\lambda)$ is given by $$\frac{d\epsilon}{d\lambda}=m\frac{\dot a}{a}V^\mu V_\mu.$$ Consider two types of particle; a massless particle and a stationary massive particle. The massless particle has $V^\mu V_\mu=0$ therefore $$ \begin{eqnarray} \frac{d\epsilon_{\rm massless}}{d\lambda} &=& 0\\ \epsilon_{\rm massless} &=& {\rm constant.} \end{eqnarray} $$ The massive particle has $V^\mu V_\mu=1$ therefore $$ \begin{eqnarray} \frac{d\epsilon_{\rm massive}}{d\tau} &=& m\frac{da/d\eta}{a}\\ &=& m\frac{da/d\tau\ d\tau/d\eta}{a}\\ &=& m\frac{da}{d\tau}\\ \epsilon_{\rm massive} &=& ma. \end{eqnarray} $$ Thus the constant energy $\epsilon_{\rm massless}$ is the energy of a massless particle as it propagates into the future with respect to the stationary observer fixed at the origin at $\eta=0$. The energy $\epsilon_{\rm massive}=ma$ is the energy of a stationary massive particle as it propagates into the future with respect to the stationary observer fixed at the origin at $\eta=0$.
Now, using natural units with $\hbar=c=1$, the Friedmann equation for the vacuum is given by $$\Big(\frac{da/d\tau}{a}\Big)^2\sim \frac{1}{M_{\rm pl}^2}\rho_{\rm vac}$$ where $M_{\rm pl}$ is the Planck mass. The vacuum density $\rho_{\rm vac}$ is given by $$\rho_{\rm vac} \sim P_{\rm pl}^4$$ where $P_{\rm pl}$ is the Planck momentum.
According to the stationary observer fixed at the origin $\eta=0$ $$ \begin{eqnarray} M_{\rm pl} &=& M_{\rm pl}^0\ a(\tau)\\ P_{\rm pl} &=& P_{\rm pl}^0 \end{eqnarray} $$ where $M_{\rm pl}^0$ and $P_{\rm pl}^0$ are the Planck mass and momentum measured at the origin.
Thus, using Planck units where $M_{\rm pl}^0=P_{\rm pl}^0=1$, the Friedmann equation for the vacuum becomes $$ \begin{eqnarray} \Big(\frac{da/d\tau}{a}\Big)^2 &\sim& \frac{1}{a^2}\\ a(\tau) &\sim& \tau. \end{eqnarray} $$ Thus the scale factor $a$ is a linear function of the proper time $\tau$ as measured by a co-moving observer. This is very different from the standard vacuum cosmology which expands exponentially.
In fact we can define the scale factor $a(\tau)=\tau$ so that the scale factor is simply the age of the universe in Planck units.
We can also redefine the energy scale so that by definition a co-moving observer always measures a constant Planck mass. In that case the energy density of the vacuum, $\rho_{\rm vac}$, must vary as $$\rho_{\rm vac} \sim \frac{1}{a^2} \sim \frac{1}{\tau^2}.$$
The age of the universe, $\tau$, is approximately $10^{60}$ in Planck units so that the current density of the vacuum $\rho_{\rm vac}\sim 10^{-120}$. This is consistent with observation.
