I'm taking "near ground" as meaning "the height over ground is negligible compared to the radius of the planet". That is, we can as good approximation we can assume that the radius of the geostationary orbit is the same as the radius of the equator.

tl;dr

Such a planet would likely be a dead, airless rock, but with interesting physical effects. Nice to put a space station on, but not developing life on its own.

Needed rotational speed

Let's first look at a perfectly spherical planet (ignoring for the moment that a planet under those extreme conditions won't be perfectly spherical). The relevant quantities for such a planet are its mass $M$, its radius $R$ and its angular velocity (rotational speed) $\omega$. We also assume the geostationary orbit at height $h\ll R$ above the equator (note that on earth, that condition would still be fulfilled at the top of Mount Everest, so it is not too limiting).

The condition of a circular orbit (which the geostationary orbit is) is that the centripetal acceleration equals the gravitational acceleration. The centripetal force is given by

$$a = \omega^2 (R+h)$$

and the gravitational acceleration is

$$g = \frac{G M}{(R+h)^2}$$

where $G = 6.7\cdot 10^{-11}\,\rm m^3\, kg^{-1}\, s^{-2}$ is the gravitational constant.

So the planet would have to spin at the angular velocity of

$$\omega = \sqrt{\frac{G M}{(R+h)^3}} \approx \sqrt{\frac{GM}{R^3}} \left(1 - \frac{3}{2} \frac{h}{R}\right)$$

To see what this means, let's insert a few numbers.

First, let's assume we've got a planet of earth mass, about $6.0\cdot 10^{24}\,\rm kg$ ($GM = 4.0\cdot 10^{14}\,\rm m^3/s^2$), and earth radius, about $6.4\cdot 10^6\,\rm m$. Then we have

$$\omega = 1.2\cdot 10^{-3}\,\rm s^{-1},$$

that is, you'd have one full rotation every 1.4 hours. Note that I neglected the height of the orbit here, since it would end up in the rounding error anyway.

Actually that's a lot less than I would have expected (but thinking again, the actual geostationary orbit's radius is just about 7 times the earth radius, so it should not have been that surprising).

Interestingly, if you look at the formula for omega, you see that the relevant quantity is the density of the planet. So we if we give the density of a planet as multiple of the earth density, omega scales with the square root of that number. So a planet of four times the earth's density would have twice the angular velocity, and thus have one rotation about every 42 minutes. On the other hand, a planet with only 1/4 of the earth's density would have 2.8 hours for each rotation. If you want to have an earth-length day (24 hours, neglecting the fact that you'd have to consider the sidereal instead of the solar day), the planet's density would have to be 0.34% of earth's density, or 19 kg/m^3. That's about 1/48 of the density of Styrofoam. A planet made completely of Styrofoam would therefore need to have a rotational period of 3.5 hours. (Note: It would be nice if someone cross-checked my numbers.)

The effects of such a rotational speed

OK, so what would be the effects of such a rotation on the planet's surface? Well, the two forces to consider are the effective gravitational force (that is, gravitation + centrifugal), and the Coriolis force. As before, I'll use accelerations instead of forces; to get the force onto an object just multiply with its mass.

Effective gravitation

The effects of course depend on the latitude, which I'll call $\phi$, in accordance to geographical conventions. It makes sense to split the acceleration into a vertical and a horizontal component, relative to the ground. The gravitational acceleration is, of course, always vertical and always the same (since we assume a spherical planet). The formula I've already given above (now we of course set $h=0$, since we are interested on the gravitation on the surface), but now we have to be careful about the direction: It points downwards, so we add a minus sign.

$$g = -G M/R^2$$

The absolute value of the centrifugal force depends on the distance from the rotation axis, which is

$$d = R \cos(\phi)$$

Otherwise, it's just the formula above, with $R$ replaced by $d$ (on the equator, of course we have $d=R$):

$$a = \omega^2 d = \omega^2 R \cos(\phi)$$

However, its direction is away from the axis, which means that we have to split it into a horizontal and a vertical component. The horizontal component is $a\sin(\phi)$, and the vertical component is $a\cos(\phi)$.

Putting everything together, we get the total vertical acceleration

$$g_{\text{eff}} = -\frac{G M}{R^2} + \omega^2 R \cos^2(\phi)$$

or, after inserting the "geostationary equator condition":

$$g_{\text{eff}} = \frac{G M}{R^2} \left(1 - \cos^2 \phi \left(1 - \frac{3}{2} \frac{h}{R}\right)^2\right) \approx \frac{G M}{R^2} \sin^2 \phi - 3 \cos^2 \phi = g \sin^2(\phi) - 3 \frac{h}{R} \cos(\phi)$$

This is exactly as expected, you're heaviest on the pole (where the rotation has no effect), and lightest at the equator (and for $h=0$, you'd be weightless at the equator).

And the force toward the equator is

$$a_{\text{eff}} = \omega^2 R \cos(\phi) \sin(\phi) \approx \frac{1}{2} \frac{G M}{R^2} \sin(2 \phi)) \left(1 - \frac{3}{2} \frac{h}{R}\right)^2 \approx \frac{g}{2} \sin(2 \phi) \left(1 - 3 \frac{h}{R}\right)$$

Note that this force is zero both at the equator and at the poles, and maximal at a latitude of 45°. At that latitude it would be half of the polar gravitation, so it would be a quite strong force. Indeed, at that latitude, the horizontal floor would have an apparent tilt of about 27°.

Coriolis force

The Coriolis force is velocity-dependent. It is the force which is responsible for the rotation of air around high/low pressure regions (and thus also in part responsible for things like hurricanes).

The Coriolis force is always perpendicular both to the axis of rotation, and to the direction of movement. Therefore we now not only have to consider the position where we are, but also the direction we are running.

I'll define the cardinal directions as on the earth: The sun rises in the east and settles in the west. The poles are in the north and south. This means that the angular momentum vector points to the north. The formula for the Coriolis acceleration is

$$\vec a_C = 2\, \vec v \times \vec\omega.$$

The horizontal direction of the Coriolis force is to the right on the northern hemisphere, and to the left on the southern hemisphere. So for example if you are running towards the closest pole (to the north on the northern hemisphere, or to the south on the southern hemisphere), the force will push you in east direction.

The most interesting part is the vertical component, which will become relevant on the equator when you're running left or right. When running on the equator in east or west direction, all of the Coriolis force is vertical; you'll get

$$a_c = 2 v \omega = 2 \frac{v}{V} R \omega^2 \approx 2 \frac{v}{V} g \left(1-\frac{3}{2} \frac{h}{R}\right) = \frac{v}{V} g \left(2 - 3 \frac{h}{R}\right).$$

where I've introduced the equatorial speed $V = R \omega$. Note that when running to the east, the force will go downwards (make you more heavy), while when running to the west, it will go upwards (make you lighter).

Compare with the effective force on the equator (see above):

$$g_{\text{eff}} = -3 g \frac{h}{R}$$

So you get an effective upwards force if you run eastwards and $a_c > g_{\text{eff}}$, that is,

$$\frac{v}{V} > \frac{3 \frac{h}{R}}{2 - 3 \frac{h}{R}} \approx \frac{3}{2} \frac{h}{R}$$

Let's calculate that with earth mass/radius, and a geostationary orbit at 8000 meter height (about Mount Everest height):

$$V = R \omega \approx \sqrt{\frac{G M}{R}} = 7.9\,\rm km/s$$

$$\implies v > 0.015\,\mathrm{km/s} = 53\,\rm km/h.$$

That's slightly above the allowed maximum driving speed inside a settlement in Germany. It's definitely much below what cars are able to do.

Would such a planet be able to develop life?

Given that above the equator there's a "gravitational leak" due to the centrifugal force, I'd not expect that planet to be able to hold an atmosphere. So if there were life on such a planet, it would certainly not be on the surface. Without much of an atmosphere, I guess also water would evaporate quite quickly, so I'd expect the planet to be mostly a dead rock. Without air, there would, of course, also not much of a climate.

What advantage would such a planet have for colonization?

Despite the disadvantage of having an effectively space-like environment, such a planet could have the advantage that you have very low launch requirements, so it would be relatively cheap to get onto/off the planet. For a space station (and possibly mining), that would be ideal.