Almost -- clocks at higher altitudes run *faster* since effect of gravitational time dilation which slows clocks closer to the Earth outweighs the special relativistic tendency to make moving clocks slow.

Not necessarily, the difference in clock rate is given by the expression:

$$d\tau_E-d\tau_S \approx \frac{dt}{2}[r_{sch}(\frac{1}{R_S}-\frac{1}{R_E})+\frac{v_S^2-v_E^2}{c^2}]$$

where:

$$d\tau_E$$=period of ground based clock

$$d\tau_S$$=period of satellite based clock

$$R_E$$=Earth radius

$$R_S$$=satellite orbit radius (assume circular orbit)

$$v_E$$=Earth tangential speed

$$v_S$$=satellite tangential speed

$$r_{sch}$$=Earth Schwarzschild radius (approx 9mm)

$$dt$$=coordinate time interval

If the satellite is geosynchronous, then:

$$d\tau_E-d\tau_S \approx \frac{dt}{2}(R_S-R_E)[(R_S+R_E)\frac{\omega^2}{c^2}-\frac{r_{sch}}{R_SR_E}]$$

The expression $$r_{sch}(\frac{1}{R_S}-\frac{1}{R_E})+\frac{v_S^2-v_E^2}{c^2}$$ can be positive, zero or negative, it all depends on the satellite orbital radius and its speed compared to the speed of the Earth.

For example, if $$v_S=\sqrt{v_E^2+c^2r_{sch}(1/R_E-1/R_S)}$$, the ground clock and the satellite clock run at the same exact rate. Granted, this is a very large speed due to the contribution of the $$c^2r_{sch}(1/R_E-1/R_S)$$.

Generally, satellites move around 8km/s, the term $$c^2r_{sch}(1/R_E-1/R_S)$$ is of the order of 30-40km/s, so, it is not impossible.

SR should not be invoked in solving this, this is a pure GR application.