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

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

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 $N=448.$