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

Given the function $$ f: (0, \infty) \to \mathbb{R} $$ is twice differentiable, and for some $$ a \neq 0 $$, the equation $$ \int_{0}^{1} f(\lambda x) d\lambda = a f(x) $$ holds, with $$ f(1) = 1 $$ and $$ f(16) = \frac{1}{8} $$. We need to find $$ 16 - f'\left( \frac{1}{16} \right) $$.

First, simplify the integral equation. Substitute $$ u = \lambda x $$, so $$ du = x d\lambda $$ and $$ d\lambda = \frac{du}{x} $$. When $$ \lambda = 0 $$, $$ u = 0 $$; when $$ \lambda = 1 $$, $$ u = x $$. Thus,
$$ \int_{0}^{1} f(\lambda x) d\lambda = \int_{0}^{x} f(u) \cdot \frac{du}{x} = \frac{1}{x} \int_{0}^{x} f(u) du $$
So the equation becomes:
$$ \frac{1}{x} \int_{0}^{x} f(u) du = a f(x) $$
Multiply both sides by $$ x $$:
$$ \int_{0}^{x} f(u) du = a x f(x) \quad \text{(1)} $$

Differentiate both sides of equation (1) with respect to $$ x $$. The left side, by the Fundamental Theorem of Calculus, is $$ f(x) $$. The right side, using the product rule, is $$ a \left[ f(x) + x f'(x) \right] $$. So,
$$ f(x) = a f(x) + a x f'(x) $$
Rearrange terms:
$$ f(x) - a f(x) = a x f'(x) $$
$$ f(x)(1 - a) = a x f'(x) $$
Thus,
$$ f'(x) = \frac{1 - a}{a} \cdot \frac{f(x)}{x} \quad \text{(2)} $$

Equation (2) is separable. Write it as:
$$ \frac{f'(x)}{f(x)} = \frac{1 - a}{a} \cdot \frac{1}{x} $$
Integrate both sides with respect to $$ x $$:
$$ \int \frac{f'(x)}{f(x)} dx = \frac{1 - a}{a} \int \frac{1}{x} dx $$
$$ \ln |f(x)| = \frac{1 - a}{a} \ln |x| + C $$
Since $$ x > 0 $$ and $$ f(x) $$ is defined for positive $$ x $$, drop the absolute values:
$$ \ln f(x) = \frac{1 - a}{a} \ln x + C $$
Exponentiate both sides:
$$ f(x) = e^C \cdot x^{\frac{1 - a}{a}} $$
Let $$ k = e^C $$, so
$$ f(x) = k x^{\frac{1 - a}{a}} \quad \text{(3)} $$

Use the given conditions. First, $$ f(1) = 1 $$:
$$ k \cdot (1)^{\frac{1 - a}{a}} = k \cdot 1 = k = 1 $$
Thus, $$ k = 1 $$, and
$$ f(x) = x^{\frac{1 - a}{a}} \quad \text{(4)} $$

Second, $$ f(16) = \frac{1}{8} $$:
$$ 16^{\frac{1 - a}{a}} = \frac{1}{8} $$
Express in terms of base 2: $$ 16 = 2^4 $$ and $$ \frac{1}{8} = 2^{-3} $$, so
$$ \left(2^4\right)^{\frac{1 - a}{a}} = 2^{-3} $$
$$ 2^{4 \cdot \frac{1 - a}{a}} = 2^{-3} $$
Equate exponents:
$$ 4 \cdot \frac{1 - a}{a} = -3 \quad \text{(5)} $$
Solve for $$ a $$:
$$ 4(1 - a) = -3a $$
$$ 4 - 4a = -3a $$
$$ 4 = 4a - 3a $$
$$ 4 = a $$
So $$ a = 4 $$.

Substitute $$ a = 4 $$ into equation (4):
$$ f(x) = x^{\frac{1 - 4}{4}} = x^{-\frac{3}{4}} $$

Now find the derivative:
$$ f(x) = x^{-\frac{3}{4}} $$
$$ f'(x) = -\frac{3}{4} x^{-\frac{3}{4} - 1} = -\frac{3}{4} x^{-\frac{7}{4}} $$

Evaluate at $$ x = \frac{1}{16} $$:
$$ f'\left( \frac{1}{16} \right) = -\frac{3}{4} \left( \frac{1}{16} \right)^{-\frac{7}{4}} $$
Note that $$ \left( \frac{1}{16} \right)^{-\frac{7}{4}} = 16^{\frac{7}{4}} $$, and $$ 16 = 2^4 $$, so
$$ 16^{\frac{7}{4}} = (2^4)^{\frac{7}{4}} = 2^{4 \cdot \frac{7}{4}} = 2^7 = 128 $$
Thus,
$$ f'\left( \frac{1}{16} \right) = -\frac{3}{4} \cdot 128 = -\frac{384}{4} = -96 $$

Finally, compute:
$$ 16 - f'\left( \frac{1}{16} \right) = 16 - (-96) = 16 + 96 = 112 $$

The answer is 112.