Short answer: $\lambda = z \mu F$ is correct. The link provided by @Loong confirms this (page 73 of the document). To add to that I have provided you with a short derivation.
Prerequisite reading (so to speak) for definitions of a few terms/notation would be my answers to this question, and this question
Let me establish an important distinction (or definition) which can (possibly) be a source of confusion. The two relevant equations from my previous answer to the first question are:
$$\Lambda _{m}=\Lambda _{m}^{0}-K{\sqrt {c}}$$
and,
$$ \Lambda_{m}^{0}= \nu_{+}\lambda_{+}^{0}+\nu _{-}\lambda _{-}^{0}$$
Before I begin my discussion, I would like to note that "$\Lambda_m$" (uppercase ) is the molar conductivity of the solution, and "$\lambda$" is the molar conductivity of the ion (ionic conductivity). The second equation has a "special name", that I did not use in my previous answer-"Kohlrausch's law of independent migration of ions".
In general for dilute solutions of strong electrolytes, molar conductivity can be decomposed into contributions from the different ions in solution (this is the spirit behind the previously quoted equation):
$$ \Lambda_m^0 = \Sigma \nu_i \lambda^0_i$$
Now, let's attempt to relate the ionic conductivity to the ionic mobility (it clearly makes sense to do so). So I am assuming we have a dilute solution of completely disassociated strong electrolyte with molar concentration $c$. Each formula unit of this electrolyte gives rise to $\nu_+$ cations of charge $z_+e$ and $\nu_-$ anions of charge $z_-e$.
The concentration of each type of ion is therefore, $\nu c$ ($\nu$ could either mean $\nu_+$, or $\nu_-$ to avoid unnecessary notational complexity).
the number density of each type of ions is:
$$\text{No. density} = \nu c N_\mathrm{A}$$
Now picture a small window of area $A$ in your solution. The number of ions that can pass through said window in a time interval $\Delta t$ is equal to the number of ions within the distance $s \Delta t$ (where $s$ is the drift speed of the ions)
$s \Delta t A$, happens to define a volume element and the number of ions in this interval is given by: $ V \times \text{no. density}$, i.e
$$ s \Delta t A \nu c N_\mathrm{A}$$
Thus, the flux of ions is
$$J_\mathrm{(ions)} = \frac{s \Delta t A \nu c N_\mathrm{A}}{ \Delta t A} = s\nu c N_\mathrm{A}$$
remember flux always has dimensions of "something" per area per time
Flux of charge is $$J_\mathrm{(charge)} = ze s\nu c N_A = z s \nu c F $$
again, I am omitting the +/- signs for the sake of clarity. Also, note $F := e \times N_\mathrm{A}$
Now, drift speed is $s = \mu E$, so the flux of charge (henceforth, simply J) is
$$J = z \mu \nu c F E$$
The current is charge flux times area, so
$$I = (z \mu \nu c F E)A$$
Additionally, $E = \frac{\Delta \phi}{l}$ (i.e the potential gradient)
$$I = (z \mu \nu c FA) \frac{\Delta \phi}{l} \tag{I}$$
Using Ohm's Law (i.e $\Delta \phi = IR$}, one obtains
$$I = \frac{\Delta \phi}{R} = G \Delta \phi = \frac{\kappa A \Delta \phi}{l} \tag{II}$$
Comparing the (I) and (II) one is lead to $$ \kappa = z \mu \nu cF$$
This is the ionic conductivity. We need to divide by concentration of ions to the get the molar quantity. The concentration of ions however is not $c$, but rather $\nu c$ and our final result is
$$\lambda = z \mu F$$
