Answer: 4

Small Hint:

Every point offers exactly two moves; call a move $X$ if it goes counterclockwise or inward and $Y$ if clockwise or outward, so paths correspond to $XY$-strings

Big Hint:

The bug is back at $A$ exactly when the last move is an $X$ and the number of $X$s exceeds the number of $Y$s by a multiple of $5$

Solution:

From any inner point the bug has exactly two moves, counterclockwise along the inner circle or outward along a spoke; from any outer point it has exactly two, clockwise along the outer circle or inward along a spoke. Call a move $X$ if it is counterclockwise or inward and $Y$ if it is clockwise or outward. Then every string in $\{X, Y\}^{15}$ describes exactly one $15$-step path from $A.$

A step arrives on the inner circle exactly when it is an $X,$ so the path ends on the inner circle exactly when its last move is an $X;$ in that case the numbers of inward and outward moves are equal. Measuring angular position in fifths of a turn (counterclockwise $+1,$ clockwise $-1,$ spokes $0$), the path returns to $A$ exactly when it ends on the inner circle and the net rotation is a multiple of $5,$ that is, when the last move is $X$ and $\#X - \#Y \equiv 0 \pmod 5.$ With $15$ moves this means the number of $X$s is $5,$ $10,$ or $15.$

Fixing the last move as $X,$ the first $14$ moves contain $4,$ $9,$ or $14$ $X$s, so $$\begin{aligned} n &= \binom{14}{4} + \binom{14}{9} + \binom{14}{14} \\ &= 1001 + 2002 + 1 = 3004, \end{aligned}$$ and the remainder is $4.$