Constant turn radius with Δ airspeed and bank
You are right on all accounts.

In a steady-state turn, from circular motion, we have, with $F_c$ being centripetal force, $V$ being airspeed in zero wind, $R$ being the radius of turn, and $m$ being the aircraft mass:

$$F_c=m\frac{V^2}{R}$$

The centripetal acceleration in a coordinated turn is created from the banked lift vector ($L$), therefore, with $\phi$ being the bank angle:

$$L\sin\phi=m\frac{V^2}{R}$$

Therefore, the radius of turn is:

$$R=\frac{V^2}{g\tan\phi}$$

and the rate of turn is:

$$\dot{\psi}=\frac{g\tan\phi}{V}$$

As speed increases, you would need an ever larger bank angle in order to achieve the same rate of turn. Of course, there's a limit to how much lift you can get out of the wing at a given speed, especially at higher Mach when transonic buffet becomes the limiting factor.