1. Given functional equation

We are given $\int_0^1 f(\lambda x)\,d\lambda = a f(x), \qquad x>0,$ with $a\neq 0$, and $f(1)=1,\qquad f(16)=\frac18.$

We need to find $16-f'\left(\frac1{16}\right).$

2. Rewrite the integral

Let $t=\lambda x$. Then $d\lambda=\frac{dt}{x}, \qquad \lambda=0\to t=0, \quad \lambda=1\to t=x.$ So $\int_0^1 f(\lambda x)\,d\lambda = \frac1x\int_0^x f(t)\,dt.$

Hence the given equation becomes $\frac1x\int_0^x f(t)\,dt = a f(x).$ Multiplying by $x$, $\int_0^x f(t)\,dt = a x f(x).$

3. Differentiate to get a differential equation

Differentiate both sides with respect to $x$: $f(x)=a\big(f(x)+x f'(x)\big).$ So $a x f'(x)+(a-1)f(x)=0.$

Rearrange: $x f'(x)+\frac{a-1}{a}f(x)=0.$

This is a first-order differential equation: $\frac{f'(x)}{f(x)}=-\frac{a-1}{a}\cdot \frac1x.$

Integrating, $\ln f(x)= -\frac{a-1}{a}\ln x + C,$ so $f(x)=C x^{-(a-1)/a}.$

Let $m=-\frac{a-1}{a}=\frac1a-1.$ Then $f(x)=C x^m.$

4. Use the given values to determine $f$

Since $f(1)=1$, $1=C\cdot 1^m \implies C=1.$ Therefore $f(x)=x^m.$

Also, $f(16)=16^m=\frac18=2^{-3}.$ Now $16=2^4$, so $(2^4)^m=2^{-3} \implies 2^{4m}=2^{-3} \implies 4m=-3.$ Thus $m=-\frac34.$

Hence $f(x)=x^{-3/4}.$

5. Differentiate $f$

$f'(x)=-\frac34 x^{-7/4}.$

Now evaluate at $x=\frac1{16}$: $f'\left(\frac1{16}\right)=-\frac34\left(\frac1{16}\right)^{-7/4}.$ Since $\left(\frac1{16}\right)^{-7/4}=16^{7/4}=(2^4)^{7/4}=2^7=128,$ we get $f'\left(\frac1{16}\right)=-\frac34\cdot 128=-96.$

6. Compute the required value

$16-f'\left(\frac1{16}\right)=16-(-96)=112.$

7. Comparison with stored answer

Our derived answer is $\boxed{112}.$ This matches the stored correct answer.