Here is the math:
##\left(\dfrac{ds}{d\tau}\right)^2=\left(\dfrac{c \, dt}{d\tau}\right)^2 - \left(\dfrac{d \vec x}{d\tau}\right)^2##,
where ##ds## is the (invariant) infinitesimal spacetime interval, ##d\tau## is the (invariant) infinitesimal proper time interval, ##dt## is the (frame-dependent) infinitesimal coordinate time interval, and ##d \vec x## is the (frame-dependent) infinitesimal displacement.
In plain English, the left side is the (squared) "speed through spacetime" of something with mass. All observers agree on its value. The right side has two terms whose value depends on an observer's velocity relative to the massive object in question: the (squared) "speed through time" and the (squared) "proper velocity."
But this expression can be simplified. First, ##ds/d\tau = c##. Second, ##dt/d\tau = 1/\sqrt{1-v^2/c^2} = \gamma##. Third, ##d \vec x / d \tau = \gamma \vec v## (where ##\vec v## is normal velocity). So we have:
##c^2=(\gamma c)^2 - (\gamma \vec v)^2##.
This means that everything with mass always moves through spacetime with a "speed" of ##c##, where "speed" means the proper-time-derivative of the spacetime interval. As ##v## increases, both terms on the right side increase, and the first of them (the time contribution) is always the bigger of the two. So it's not quite true that "speed through time" decreases as ##v## increases.
What is true is that an object at rest has all of its "motion" through spacetime directed forward in time, whereas moving observers would say that the object moves through both space and time. All observers agree, however, that the difference of the squares of the space and time contributions equals ##c^2##, and that the object moves through spacetime at the ##\tau##-rate of ##c##.
