Wave drag at high altitudes is high or low? Density decreases with increasing altitude. Consequently, Reynolds number decreases and boundary layer thickness increases, so friction drag and pressure drag increase.
What about wave drag? Does wave drag increase or decrease with increasing altitude?
 A: Well, first let's set things straight. The Reynolds number is defined as
$$Re = \frac{\rho uL}{\mu},$$
where $\rho$ is the density, $u$ is the flow velocity, $L$ is a typical length scale of the flow, and $\mu$ is the dynamic viscosity. Reducing the density, while keeping flow velocity and temperature (thus viscosity) constant, leads to a reduction in the Reynolds number. However, this doesn't mean at all that friction drag increases, quite the opposite in fact. Indeed, the friction drag coefficient increases, since it is inversely proportional to the Reynolds number, and for an incompressible, turbulent flow over a flat plate is given by
$$C_f = \frac{0.074}{Re^{1/5}}.$$
However, friction drag itself is also a function of density:
$$F = \frac{1}{2}\rho A u^2 C_f,$$
where $A$ is a reference area. Since density is linearly proportional to $Re$, we have $F \propto Re^{4/5}$, and will thus decrease with the decrease of Reynolds number.
About wave drag: this form of drag is caused by the sharp increase in pressure behind shockwaves surrounding a body propagating at supersonic speed. Pressure ratio across a shockwave at a given angle depends only on the Mach number of the shock. Thus, if an aircraft keeps moving at a constant Mach number while ascending in the atmosphere, the pressure ratio across the shocks it generates would remain constant, and I would expect it to face lower pressure drag, since the pressure upstream of the shock (and thus also behind the shock) decreases with altitude.
