Yes, for a circular orbit at about $1.50$ Earth radii, the gravitational and special-relativistic time dilations cancel. GPS is modeled with the static weak-field metric (in units of $c = 1$):
$$-d\tau^2 = -(1+2\Phi)dt^2 + (1-2\Phi)\underbrace{(dx^2+dy^2+dz^2)}_{dS^2},$$
so we can approximate
$$d\tau = dt\sqrt{1+2\Phi-(1-2\Phi)\frac{dS^2}{dt^2}}\approx\left[1+\Phi-\frac{v^2}{2}\right]dt,$$
with corrections due to gravitational time dilation and special-relativistic time dilation additively to this order, since the latter gives a factor of $1/\gamma = 1 - \frac{v^2}{2} + \mathscr{O}(v^4)$, while stationary gravitational time dilation would have a factor of $\sqrt{1+2\Phi} = 1 + \Phi + \mathscr{O}(\Phi^2)$.
To the above order of approximation, every clock on the geoid ticks at the same rate because the geoid is an equipotential surface of the sum of the gravitational and centrifugal potentials (this is sometimes called gravity potential). This common factor happens to be $1 - \alpha$, where $\alpha \approx 6.9692\times10^{-10}$.
Since the metric is static, $\partial_t$ is a Killing field producing conservation of orbital specific energy $\epsilon = (1+2\Phi)\frac{dt}{d\tau}$. If we want the satellite to have the same time dilation as the clock on the geoid, then by substitution, $\Phi$ must be a constant, and consequently so must be $v$. In other words, it is not possible to have an orbit that plays off changes in gravitational potential against changes in velocity so as to keep the overall time dilation the same.
Now, if we approximate the Earth as spherically symmetric with gravitational potential $\Phi = -\frac{\mu}{r}$, then it is clear that only circular orbits need be considered, and the same tick rate condition $-\Phi + \frac{v^2}{2} = \alpha$ gives $\alpha = \frac{3}{2}\frac{\mu}{r}$, or $1.50$ Earth radii.
Actually, the gravitational potential in the GPS model is taken to be the monopole and quadrupole terms:
$$\Phi = -\frac{\mu}{r}\left[1-J_2\left(\frac{a_1}{r}\right)^2P_2(\cos\theta)\right],$$
where $\mu = GM_\text{E} = 3.986004418(9)\times 10^{14}\,\text{m}^3/\text{s}^2$ is the standard gravitational parameter, $J_2 = 1.0826300\times10^{-3}$ is the quadrupole moment coefficient, $a_1 = 6.3781370\times10^6\,\text{m}$ is the equatorial radius, $\theta$ is the usual polar angle, and $P_2(x) = \frac{1}{2}(3x^2-1)$ is a Legendre polynomial.
This situation is a bit harder, but you should still be able to get an orbit that gives close agreement in the equatorial plane.
