Answer
$ 1.16\;\rm mm/s$
Work Step by Step
We know that the lateral magnification is given by
$$m=-\dfrac{s'}{s}\tag 1$$
We also know that the longitudinal magnification is given by
$$M= \dfrac{ds'}{ds}\tag 2$$
See problem 82 in your textbook in this chapter, it refers to the same $M$.
For the thin lens,
$$\dfrac{1}{s}+\dfrac{1}{s'}=\dfrac{1}{f}$$
Solving for $s'$,
$$s'=\left[ \dfrac{1}{f}-\dfrac{1}{s} \right]^{-1}=\left[ \dfrac{s-f}{fs} \right]^{-1}$$
$$s'=\dfrac{fs}{s-f} \tag 3$$
Plugging $s'$ from (3) into (1),
$$m=-\dfrac{ 1}{s}\cdot \dfrac{fs}{s-f}$$
$$m=-\dfrac{ f}{s-f} \tag 4 $$
Plugging $s'$ from (3) into (2),
$$M= \dfrac{d }{ds}\left[\dfrac{fs}{s-f}\right]=\dfrac{f(s-f)-fs}{(s-f)^2} $$
$$M= \dfrac{fs-f^2-fs}{(s-f)^2} =-\dfrac{f^2}{(s-f)^2}$$
$$M= -\left[\dfrac{f }{(s-f)}\right]^2=-m^2\tag{From (4)}$$
$$\boxed{M= -m^2}\tag 5$$
Now we are given that the speed of the object is 5 m/s which means that
$$v_{\rm object}=\dfrac{ds}{dt}=5\;\rm m/s\tag 6$$
From (2) and (5),
$$M=\dfrac{ds'}{ds}=-m^2\tag 7$$
Now we need to find $v_{\rm image}=ds'/dt$,
$$v_{\rm image}=\dfrac{ds'}{dt}=\dfrac{ds'}{ds}\dfrac{ds}{dt}$$
Plugging from (6) and (7),
$$v_{\rm image}=-m^2(5)=-5m^2$$
Plugging $m$ from (4),
$$v_{\rm image}= -5\left[ -\dfrac{ f}{s-f} \right]^2$$
when the object (player) is at 10 m in front of the lens,
$$v_{\rm image}= -5\left[ -\dfrac{ 150\times 10^{-3}}{10-(150\times 10^{-3})} \right]^2$$
$$v_{\rm image}= \color{red}{\bf -1.16}\;\rm mm/s$$
The negative sign means that the image is moving toward the lens while the object is moving away from the lens.
