Answer
$$ F_{n,pole}\gt F_{n,equator} $$
Work Step by Step
We know that the gravitational force exerted on some object of mass $m$ is given by
$$F_G=\dfrac{GmM_E}{R_E^2}$$
where $G$ is gravitational constant, $M_E$ is the mass of Earth, and $R_E$ is Earth's radius.
We know that the Earth's radius at the equator is greater than its radius on the north pole $(R_{E,equtor}\gt R_{E, pole})$.
Thus,
$$\boxed{F_{G,pole}\gt F_{G,equtor}} $$
This means that the reading of the scale on the north pole is greater than that of the equator.
Noting that we ignored the rotational motion of the earth here. Now let's assume that Earth's radius is the same in both cases.
Let's work with involving Earth's rotation.
At the north pole, the normal force (which is the reading of the scale) exerted on our man is just equal to its own weight, see the figures below, and is given by
$$\sum F_r= F_{n,pole}-mg=ma_r=m(0)=0$$
$$F_{n,pole}=mg=75\times 9.8=\color{red}{\bf 735}\;\rm N\tag 1$$
And at the equator,
$$\sum F_r=mg-F_{n,equator} =ma_r=m\dfrac{v^2}{R_E}$$
Thus,
$$F_{n,equator}=mg-\dfrac{mv^2}{R_E} $$
where $v=\dfrac{2\pi R_E}{T}$ and $T=24$ h which is the periodic time for one full rotation of the earth around its axis.
$$F_{n,equator}=mg-\dfrac{m(2\pi)^2R_E^{ \color{red}{\bf\not}2}}{T^2 \color{red}{\bf\not}R_E} $$
$$F_{n,equator}=mg-\dfrac{4\pi^2 m R_E }{T^2 } $$
Plugging the known;
$$F_{n,equator}=(75\times 9.8)-\dfrac{4\pi^2 (75) (6.37\times 10^6)}{(24\times 60^2)^2 }$$
$$F_{n,equator}= \color{red}{\bf 732.5}\;\rm N\tag 2$$
Hence,
$$\boxed{F_{n,pole}\gt F_{n,equator}}$$
Therefore the reading of the scale on the north pole is greater than that of the equator for two main reasons, the change in the radius and the rotation of the Earth around its axis.
