The answer to your question depends on the coil. Assuming the coil is operated at a frequency that is fixed and reasonable for its construction, and ignoring winding resistance, coils without magnetic cores ("air core inductors") have a constant impedance because the permeability of the core is unchanged regardless of the current and the resulting field strength.
Most inductors have ferromagnetic cores whose permeability varies with field strength, so your inductance will decrease as the current increases. If you look at the magnetization curve (B-H curve) for a core material you can see that the curve flattens out as core saturates. Coils are designed to operate in the area around the origin so that the saturation effect is not as pronounced, but there is some small effect even in this area.
In addition, losses for cores occur ("hysteresis losses") when the direction of magnetization is reversed. So, a well-designed ferromagnetic inductor has a small variance in inductance with current that is unavoidable, as you have suggested. The impedance will be lower (the inductance will drop) slightly as the current increases. If the core is saturated, the inductance will drop dramatically.
My stab at it (revised). The original block quote :
If you take a loop of superconducting wire, and apply 1V to this wire during 1s, then the magnetic flux inside this loop will have changed by 1Wb.
With qualifications that this is independent of size, shape. material ... but with no qualification about the number of turns. This leads to:
Wb = V * s ... eq1
It says nothing about the current flowing in the turn (or turns) and leaves unanswered whether an N turn coil obeys
Wb = V * s ... eq1a
or
Wb = V * s * N ... eq1b
or even
Wb = V * s / N ... eq1c
Note the definition of Weber
The weber is the magnetic flux that, linking a circuit of one turn, would produce in it an electromotive force of 1 volt if it were reduced to zero at a uniform rate in 1 second
(yes from Wiki but that links to a primary reference) so it is the flux related to 1 V-s explicitly in a single turn. A crucial difference of phrasing absent from the linked page...
A second turn in the same field would be an independent voltage source.
This brings the definition in line with eq1c because 1 Weber is the flux related to 1V-S per turn.
So my (revised!) understanding of the original quote is
If you take a loop of superconducting wire, and apply 1V per turn to this wire during 1s, then the magnetic flux inside this loop will have changed by 1Wb.
This supports Andy's understanding of Faraday's Law expressed in the question - to keep the rate of change of flux constant, you need to keep the voltage per turn constant. Alternatively, if you halve the voltage per turn you will indeed halve the rate of change of flux.
It also leads to the modification in Eq1 of the linked webpage. Which then leads logically to his final equation
H = Wb * turns / A
or
Wb = H * A / turns
This originally made me suspicious, because one normally sees flux as proportional to ampere-turns, so amperes/turn looked ... unfamiliar. The reason is that the inductance already contains a turns-squared term :
L = Al * n^2 (where Al is called "specific inductance" and is a constant for a particular geometry and material)
H = Al * turns^2
Substituting for inductance brings us back to the familiar ampere-turns
Wb = Al * A * turns
which is a more convenient form for some purposes in inductor design.
Best Answer
Let's review the formulas that apply:
Weber-Turns (flux linkage): $$ \lambda=N\Phi $$
Flux: $$ \Phi=\frac {Ni} {\Re}$$
where \$\Re\$ is the reluctance, that depends on the coil's core material and geometry.
Voltage (emf)
$$ V=\frac {d\lambda}{dt}=\frac {N^2} {\Re} \frac {di}{dt}=L\frac {di}{dt}$$
The problem here is that it is not specified how the coil is excited.
Coil excited by a sinewave current
If it is excited by a sinewave current, then it is rather clear that lower inductance, implies lower flux linkage, lower emf. About the flux it is given by \$\Phi=\frac {Ni} {\Re}=\frac L N i\$. So reducing L alone doesn't give an indication of what \$\Phi\$ will do. For example, N can be reduced in more proportion than L if one choose a different coil material for example. Therefore \$\Phi\$ can be bigger or lesser than before.
Coil excited by a voltage
In contrast, if the coil excited by a voltage, then the flux linkage is given by the voltage alone (because \$V=\frac {d\lambda}{dt}\$, linkage is the voltage integration), and so doesn't depends on the inductance. Also the flux doesn't depends really on the inductance in this case, but just on the number of turns. (And because if wished you can have a big inductance with few turns by selecting a different core, then you might have a big or smaller flux by reducing L). Finally, again if excited by a voltage, the emf will be equal to it, so doesn't depends on the inductance.
Actually whatever excitation you consider, the first 3 options are wrong.
The last option asks about a steady current, so one can assume it is excited by a voltage in series with a resistor.