6. Let the real number $a$ and the function $f(x)$ satisfy: $f(1)=1$, and for any $x \in \mathbf{R} \backslash\{0\}$, $f\left(\frac{1}{x}\right)=a f(x)-x$. Let $A=\{x|x \in \mathbf{R} \backslash\{0\}| f,(x) \mid \leqslant 1\}$. Then the range of the function $g(x)=f(a x) f\left(\frac{1}{x}\right)(x \in A)$ is $\qquad$
6. $\left[\frac{25}{18}, \frac{3}{2}\right]$.

In $f\left(\frac{1}{x}\right)=a f(x)-x$, let $x=1$, we get
$$
1=a-1 \Rightarrow a=2 \text {. }
$$

Thus, for any $x \in \mathbf{R} \backslash\{0\}$, we have
$$
f\left(\frac{1}{x}\right)=2 f(x)-x \text {. }
$$

Substituting $\frac{1}{x}$ for $x$ gives
$$
\begin{array}{l} 
f(x)=2 f\left(\frac{1}{x}\right)-\frac{1}{x} . \\
\text { Then } f(x)=\frac{1}{3}\left(2 x+\frac{1}{x}\right) \\
\Rightarrow g(x)=f(2 x) f\left(\frac{1}{x}\right) \\
=\frac{1}{3}\left(4 x+\frac{1}{2 x}\right) \cdot \frac{1}{3}\left(\frac{2}{x}+x\right) \\
=\frac{1}{9}\left(\frac{17}{2}+4 x^{2}+\frac{1}{x^{2}}\right) .
\end{array}
$$

Notice that,
$$
\begin{array}{l}
x \in A \Leftrightarrow|f(x)| \leqslant 1 \\
\Leftrightarrow\left(\frac{1}{3}\left(2 x+\frac{1}{x}\right)\right)^{2} \leqslant 1 \\
\Leftrightarrow 4 x^{2}+\frac{1}{x^{2}} \leqslant 5 .
\end{array}
$$

Since $4 x^{2}+\frac{1}{x^{2}} \geqslant 2 \sqrt{4 x^{2} \cdot \frac{1}{x^{2}}}=4$, combined with continuity, we know that when $x \in A$, $4 x^{2}+\frac{1}{x^{2}}$ takes all values in the interval $[4,5]$.
Thus, when $x \in A$,
$$
g(x) \in\left[\frac{1}{9}\left(\frac{17}{2}+4\right), \frac{1}{9}\left(\frac{17}{2}+5\right)\right],
$$

i.e., the range of $g(x)$ is $\left[\frac{25}{18}, \frac{3}{2}\right]$.