To talk about 'the rate of time', we essentially need at least two different time coordinates. For example, this happens in special-relativistic time dilation, which is equivalent to $\mathrm{d}t'/\mathrm{d}t$ across two different inertial frames. Fortunately, we can do something similar here.

Space expands everywhere, also here. And time is inseparable from space. Does this mean that time also "expands" as in changing its pace? ... In a similar way that the expansion of space is compared relative to, well, to itself I suppose.

A spatially isotropic and homogeneous universe has the metric in the form $$\mathrm{d}s^2 = -\mathrm{d}t^2 + a^2(t)\mathrm{d}\Sigma^2\text{,}$$ where $a(t)$ is the scale factor and $\mathrm{d}\Sigma^2$ is the metric of an isotropic and homogeneous Riemannian manifold: the 'open' hyperbolic $3$-plane, the flat Euclidean $3$-space, or the 'closed' $3$-sphere (or real projective $3$-space, but that's usually not considered because it's non-orientable). If conformal factor is ever zero in the past, the cosmological time for this is conventionally chosen to be $t = 0$.

The cosmological time measures the proper time of an observer at rest relative to the bulk of the matter in the universe, so in some sense it's the most intuitive choice of a time coordinate, but like all coordinates, it's not sacred. We can, for example, define a conformal time coordinate $\eta$ such that $\mathrm{d}\eta = \mathrm{d}t/a$, in which the metric takes the form $$\mathrm{d}s^2 = a^2(\eta)\left[-\mathrm{d}\eta^2 + \mathrm{d}\Sigma^2\right]\text{,}$$ and so all of the dimensions of spacetime are affected by cosmic expansion in the same way. Therefore, I think conformal time satisfies the requirements in your question, although it is not measured by any local clock.

Is the changing rate of time also astronomically observable?

The scale factor is astronomically observable, and $\mathrm{d}\eta/\mathrm{d}t = 1/a$, so yes.

How did time behave during the radical inflation shortly after Big Bang?

The conformal time essentially uses the particle horizon as a measure of time, i.e. the furthest distance an ideal lightlike signal could have travelled since $t = 0$ in order to reach the observer. During inflation, the particle horizon rapidly expanded.

To talk about 'the rate of time', we essentially need at least two different time coordinates. For example, this happens in special-relativistic time dilation, which is equivalent to $\mathrm{d}t'/\mathrm{d}t$ across two different inertial frames. Fortunately, we can do something similar here.

Space expands everywhere, also here. And time is inseparable from space. Does this mean that time also "expands" as in changing its pace? ... In a similar way that the expansion of space is compared relative to, well, to itself I suppose.

A spatially isotropic and homogeneous universe has the metric in the form $$\mathrm{d}s^2 = -\mathrm{d}t^2 + a^2(t)\mathrm{d}\Sigma^2\text{,}$$ where $a(t)$ is the scale factor and $\mathrm{d}\Sigma^2$ is the metric of an isotropic and homogeneous Riemannian manifold: the 'open' hyperbolic $3$-plane, the flat Euclidean $3$-space, or the 'closed' $3$-sphere (or real projective $3$-space, but that's usually not considered because it's non-orientable). If the scale factor is ever zero in the past, the cosmological time for this is conventionally chosen to be $t = 0$.

The cosmological time measures the proper time of an observer at rest relative to the bulk of the matter in the universe, so in some sense it's the most intuitive choice of a time coordinate, but like all coordinates, it's not sacred. We can, for example, define a conformal time coordinate $\eta$ such that $\mathrm{d}\eta = \mathrm{d}t/a$, in which the metric takes the form $$\mathrm{d}s^2 = a^2(\eta)\left[-\mathrm{d}\eta^2 + \mathrm{d}\Sigma^2\right]\text{,}$$ and so all of the dimensions of spacetime are affected by cosmic expansion in the same way. Therefore, I think conformal time satisfies the requirements in your question, although it is not measured by any local clock.

Is the changing rate of time also astronomically observable?

The scale factor is astronomically observable, and $\mathrm{d}\eta/\mathrm{d}t = 1/a$, so yes.

How did time behave during the radical inflation shortly after Big Bang?

The conformal time essentially uses the particle horizon as a measure of time, i.e. the furthest distance from which an ideal lightlike signal could have travelled since $t = 0$ in order to reach the observer by the present time. During inflation, the particle horizon rapidly expanded.