Let $f:(0,\infty) \rightarrow \mathbf{R}$ be a twice differentiable function. If for some $\mathrm{a} \neq 0$, $\int_0^1 f(\lambda x)\,\mathrm{d}\lambda = \mathrm{a}f(x)$, $f(1) = 1$ and $f(16) = \frac{1}{8}$, then $16 - f'\left(\frac{1}{16}\right)$ is equal to \_\_\_\_ .