Answer: 448

Small Hint:

Center the $6$-by-$8$ rectangle at the origin and parameterize consecutive inner vertices on the right and top sides

Big Hint:

Use equal half-diagonals to relate the two free coordinates, then express the square of the perimeter through the diagonal and area

Solution:

Opposite vertices lie on opposite sides of the outer rectangle, so the two rectangles have the same center. For a small rotation through angle $\delta,$ a right-side contact at $(4,y)$ moves inward only if $y\delta\geq0,$ while a top-side contact at $(u,3)$ moves inward only if $u\delta\leq0.$ Thus an unstuck rectangle has opposite-signed offsets; after reflection, write its consecutive vertices as $(4,y),(-x,3),(-4,-y),(x,-3)$ with $x,y\geq0.$ Equal half-diagonals give $16+y^2=x^2+9,$ so $x^2-y^2=7.$ If its side lengths are $a,b,$ then $$\begin{aligned}a^2+b^2&=4(16+y^2),\\ab&=24+2xy.\end{aligned}$$ Therefore its perimeter $Q=2(a+b)$ satisfies $$\begin{aligned}Q^2&=4(a^2+b^2+2ab)\\&=448+16y(y+x)\geq448.\end{aligned}$$ Equality occurs at $y=0,\ x=\sqrt7,$ which gives a non-axis-aligned, hence unstuck, rectangle. Thus the minimum perimeter is $\sqrt{448}$ and $N=448.$