1. Acceleration on a smooth $45^\circ$ incline

For a perfectly smooth incline, $a_s = g\sin 45^\circ = \frac{g}{\sqrt{2}}$

If the object starts from rest and slides a distance $s$, then $s = \frac{1}{2} a_s t_s^2$ where $t_s$ is the time on the smooth plane.

2. Acceleration on a rough $45^\circ$ incline

For the rough incline, friction acts up the plane.

Normal reaction: $N = g\cos 45^\circ \cdot m$

Friction force: $f_k = \mu N = \mu mg\cos 45^\circ$

Net acceleration down the plane: $a_r = g\sin 45^\circ - \mu g\cos 45^\circ$

Since $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$, $a_r = \frac{g}{\sqrt{2}}(1-\mu)$

If $t_r$ is the time on the rough plane, then $s = \frac{1}{2} a_r t_r^2$

3. Use the given time relation

Given: $t_r = n t_s$

For the same distance $s$ from rest, time is inversely proportional to the square root of acceleration: $t \propto \frac{1}{\sqrt{a}}$

So, $\frac{t_r}{t_s} = \sqrt{\frac{a_s}{a_r}} = n$

Squaring both sides, $n^2 = \frac{a_s}{a_r}$

Substitute $a_s = \frac{g}{\sqrt{2}}$ and $a_r = \frac{g}{\sqrt{2}}(1-\mu)$: $n^2 = \frac{\frac{g}{\sqrt{2}}}{\frac{g}{\sqrt{2}}(1-\mu)} = \frac{1}{1-\mu}$

Thus, $1-\mu = \frac{1}{n^2}$

Hence, $\mu = 1 - \frac{1}{n^2}$

4. Check with options

This matches: $\boxed{\text{A: } 1 - \frac{1}{n^2}}$