$$
\begin{array}{l}
\text { 1. Let } f(x)=k x+b \text {, and let } \\
A=\{x \mid f(x)=x, x \in \mathbf{R}\}, \\
B=\{x \mid f(f(x))=x, x \in \mathbf{R}\} .
\end{array}
$$

If $A \neq B$, then in the $x O y$ plane, the figure formed by the points $(k, b)$ is $\qquad$
1. The line $x=-1$. From $f(x)=x$, we get $(1-k) x=b$.
When $k=1, b=0$, $A=\mathbf{R}$; when $k=1, b \neq 0$, $A=\varnothing$; when $k \neq 1$, $A=\left\{\frac{b}{1-k}\right\}$. From $f(f(x))=x$, we get $\left(1-k^{2}\right) x=k b+b$. When $k=1, b=0$, $B=\mathbf{R}$; when $k=1, b \neq 0$, $B=\varnothing$; when $k=-1$, $B=\mathbf{R}$; when $k \neq \pm 1$, $B=\left\{\frac{b}{1-k}\right\}$. In summary, only when $k=-1$, $A \neq B$. Therefore, the figure formed by the points $(k, b)$ is the line $x=-1$.