1. Write the relevant reduction half-reaction

For the couple $MnO_4^- / Mn^{2+}$ in acidic medium:

$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

2. Write the Nernst equation

For the reduction potential,

$E = E^\circ - \frac{0.059}{5}\log\left(\frac{[Mn^{2+}]}{[MnO_4^-][H^+]^8}\right)$

Given that the concentrations of $MnO_4^-$ and $Mn^{2+}$ remain the same, only $[H^+]$ changes. So the change in potential depends only on the $[H^+]^8$ term.

3. Find the change in potential when $[H^+]$ changes

Initially,

$[H^+]_1 = 1\,M$

Finally,

$[H^+]_2 = 10^{-4}\,M$

Using the Nernst equation, the change in reduction potential is:

$\Delta E = E_2 - E_1$

Since only $[H^+]$ changes,

$E = E^\circ - \frac{0.059}{5}\log\left(\frac{\text{constant}}{[H^+]^8}\right)$

This can be written as

$E = \text{constant} + \frac{0.059}{5}\cdot 8\log[H^+]$

Hence,

$\Delta E = \frac{8\times 0.059}{5}\left(\log[H^+]_2 - \log[H^+]_1\right)$

Now,

$\log(10^{-4}) = -4, \quad \log(1)=0$

So,

$\Delta E = \frac{8\times 0.059}{5}(-4)$

$\Delta E = -\frac{1.888}{5}$

$\Delta E = -0.3776\,V$

Thus, the magnitude of change in oxidising power is

$|\Delta E| = 0.3776\,V$

4. Match with the given form

The question states magnitude $= x \times 10^{-4}\,V$.

So,

$0.3776 = x \times 10^{-4}$

$x = 0.3776 \times 10^4 = 3776$

5. Final answer

$\boxed{x=3776}$