I don't think that I saw your post with the driver type.
This is a possible circuit. The "IN" connection needs a voltage in the 0 - 1 V range, but a potential divider can be used to get 0 - 10 V down to that.
The "OUT" connection goes to the RSET connection on the Cat4101.
You need a positive supply to the op-amp, which I hope is obvious.
The resistor needs to be 5/6ths of the value that you would use with the Cat4101 to give the maximum current that you want for the 10 V input.
The transistor is just about any NPN transistor. The op-amp needs to have a common-mode range and output that can go down to the negative rail.
The Cat4101 holds the RSET pin at about 1.2 V and holds the LED current at 400 times the current flowing out of the RSET pin. My suggested circuit controls the voltage on the resistor the same as the input voltage, so the current in the resistor is proportional to the input voltage. The resistor current comes from the emitter of the transistor and just about all of the emitter current comes from the collector, which is from the RSET pin.
So the LED current is about
400 x Vin / R * (1 - 1/hfe) where Vin is the input voltage, R is the resistor value, and hfe is transistor gain
That won't be exact, because the Cat4101 factor of 400 isn't exact, and the typical values show the factor reducing by about 10% from 100 mA to 1 A