In a jet, are Vy and Vx achieved at a constant AOA at all altitudes?
It might help to derive both speeds from first principles. We assume that the drag polar of the aircraft can be described by a parabola, like this:
$$c_D = c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$
 The symbols are:
$\kern{5mm} c_D \:\:\:$  drag coefficient
$\kern{5mm} c_{D0} \:$  zero-lift drag coefficient
$\kern{5mm} c_L \:\:\:$  lift coefficient
$\kern{5mm} \pi \:\:\:\:\:$  3.14159$\dots$
$\kern{5mm} AR \:\:$  aspect ratio of the wing
$\kern{5mm} \epsilon \:\:\:\:\:\:$  the wing's [Oswald factor][4]

Next, we describe thrust $T$ over speed $v$ with $$T = T_0·v^{n_v}$$

Now first to the maximum climb angle. This is reached when the condition $$\frac{\delta \gamma}{\delta c_L} = 0$$
holds true. No change in lift coefficient $c_L$ will improve the climb angle $\gamma$, it is only downhill from here to both sides. In order to get a grip on the climb angle, we look at the force equilibrium in steady flight at full power, assuming small values for $\gamma$:
$$sin\gamma = \gamma = \frac{v_z}{v} = \frac{T - c_D\cdot \frac{\rho}{2}\cdot v^2\cdot S_{ref}}{m\cdot g} = \frac{T_0·\left(\sqrt{\frac{2\cdot m\cdot g}{\rho\cdot c_L\cdot S_{ref}}}\right)^{n_v}}{m\cdot g} - \frac{c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}}{c_L}$$
 The symbols are:
$\kern{5mm} m\:\:\:\:$ aircraft mass
$\kern{5mm} g\:\:\:\:\;$ gravitational acceleration
$\kern{5mm} \rho\:\:\:\:\:$ air density
$\kern{5mm} v\:\:\:\:\:$ velocity
$\kern{5mm} v_z\:\:\;$ climb speed
$\kern{5mm} S_{ref} \:$  wing area

Ideally, we would also multiply the climb angle with an acceleration factor, but I leave this out here for simplicity.

Now we can derive the expression for the climb angle with respect to the lift coefficient and get
$$\frac{\delta \gamma}{\delta c_L} = -\frac{n_v}{2}·c_L^{-\frac{n_v}{2}-1}·\frac{T_0·(m·g)^{\frac{n_v}{2}-1}}{\left(\frac{\rho}{2}·S_{ref}\right)^{\frac{n_v}{2}}}+\frac{c_{D0}}{c_L^2}-\frac{1}{\pi·AR·\epsilon}$$
The general solution is $$c_{L_{{\gamma_{max}}}} = -\frac{n_v}{4}·\frac{T·\pi·AR·\epsilon}{m·g}+\sqrt{\frac{n_v^2}{16}·\left(\frac{T·\pi·AR·\epsilon}{m·g}\right)^2+c_{D0}·\pi·AR·\epsilon}$$
For jets ($n_v = 0$) the solution is quite simple, because the thrust terms are proportional to the thrust coefficient $n_v$ and disappear: $$c_{L_{{\gamma_{max}}}} = \sqrt{c_{D0}·\pi·AR·\epsilon}$$
For turbofan and propeller aircraft, we have less luck and get a much longer formula. This is the one for propellers ($n_v = -1$):
$$c_{L_{{\,\gamma_{\,max}}}} = \frac{T·\pi·AR·\epsilon}{4·m·g}+\sqrt{\left(\frac{T·\pi·AR·\epsilon}{4·m·g}\right)^2+c_{D0}·\pi·AR·\epsilon}$$
So yes, pure turbojets have an optimum lift coefficient for maximum climb angle that uses only terms which are constant over altitude. They do indeed climb steepest at a constant lift coefficient.

But the thrust-dependent optimum for other engine types hints at an altitude dependency which might affect the other optimum, that for best climb speed.

In order to find the conditions for maximum climb speed, repeat the process above with an expression where both sides are multiplied by speed:
$$v_z = \frac{T\cdot v - c_D\cdot \frac{\rho}{2}\cdot v^3\cdot S_{ref}}{m\cdot g} = \frac{T_0·\left(m\cdot g\right)^{\frac{n_v-1}{2}}}{\left(c_L\cdot\frac{\rho}{2}\cdot v^2\cdot S_{ref}\right)^{\frac{n_v+1}{2}}} - \sqrt{\frac{2\cdot m\cdot g}{\rho\cdot c_L\cdot S_{ref}}}\cdot\frac{c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}}{c_L}$$

Now the solution for turbojets becomes the more complicated one, but that needs to be the case - how else would those optima converge at altitude?
$$c_{L_{{\,n_{z_{\,max}}}}} = \sqrt{\left(\frac{T·\pi·AR·\epsilon}{2·m·g}\right)^2 + 3\cdot c_{D0}·\pi·AR·\epsilon} - \frac{T·\pi·AR·\epsilon}{2·m·g}$$

While the angle of attack for steepest climb is constant over altitude, the angle of attack for best climb speed increases as excess thrust disappears with rising altitude. Thus, the question can be answered: No.