If $\angle A$ is four times $\angle B$, and the complement of $\angle B$ is four times the complement of $\angle A$, then $\angle B=$

$ \textbf{(A)}\ 10^{\circ} \qquad\textbf{(B)}\ 12^{\circ} \qquad\textbf{(C)}\ 15^{\circ} \qquad\textbf{(D)}\ 18^{\circ} \qquad\textbf{(E)}\ 22.5^{\circ} $
1. Let $\angle A = x$ and $\angle B = y$. According to the problem, $\angle A$ is four times $\angle B$, so we have:
   \[
   x = 4y
   \]

2. The complement of an angle is $90^\circ$ minus the angle. Therefore, the complement of $\angle B$ is $90^\circ - y$ and the complement of $\angle A$ is $90^\circ - x$.

3. The problem states that the complement of $\angle B$ is four times the complement of $\angle A$. Thus, we have:
   \[
   90^\circ - y = 4(90^\circ - x)
   \]

4. Substitute $x = 4y$ into the equation:
   \[
   90^\circ - y = 4(90^\circ - 4y)
   \]

5. Simplify the right-hand side:
   \[
   90^\circ - y = 4 \cdot 90^\circ - 4 \cdot 4y
   \]
   \[
   90^\circ - y = 360^\circ - 16y
   \]

6. Rearrange the equation to isolate $y$:
   \[
   90^\circ - y = 360^\circ - 16y
   \]
   \[
   90^\circ - 360^\circ = -16y + y
   \]
   \[
   -270^\circ = -15y
   \]

7. Solve for $y$:
   \[
   y = \frac{270^\circ}{15}
   \]
   \[
   y = 18^\circ
   \]

The final answer is $\boxed{18^\circ}$