6. 154 Find all monotonic invertible functions \( f: R \rightarrow R \) that satisfy
$$f(x)+f^{-1}(x) \equiv 2 x, x \in R$$
[Solution] Let $f(x)$ satisfy the conditions given in the problem. Consider the function $g(x) = f(x) - x$. First, we prove that for any $k \in \mathbb{Z}^{+}$, we have
$$f(x + k g(x)) \equiv x + (k+1) g(x), \quad x \in \mathbb{R}$$

When $k=0$, the identity $f(x) \equiv x + g(x)$ holds. Assume the identity holds for some $k \in \mathbb{N}$, i.e.,
$$f(x + k g(x)) \equiv x + (k+1) g(x)$$

From this, we know that $f^{-1}(x + (k+1) g(x)) \equiv x + k g(x)$. By the condition given in the problem, for all $x \in \mathbb{R}$, we have
$$\begin{aligned}
f(x + (k+1) g(x)) \equiv & 2(x + (k+1) g(x)) - f^{-1}(x + (k+1) g(x)) \\
\equiv & 2(x + (k+1) g(x)) - (x + k g(x)) \\
\equiv & x + (k+2) g(x).
\end{aligned}$$

Thus, the identity also holds for $k+1$. This also implies that for all $k \in \mathbb{N}$, the identity
$$f^{-1}(x + k g(x)) \equiv x + (k-1) g(x), \quad x \in \mathbb{R}$$

holds. Similarly, we can prove that for all $k \in \mathbb{Z}^{+}$,
$$f^{-1}(x - (k-1) g(x)) \equiv x - k g(x), \quad x \in \mathbb{R}$$

From this, we get that for all $k \in \mathbb{N}$,
$$f(x - k g(x)) \equiv x - (k-1) g(x), \quad x \in \mathbb{R}$$

Next, note that the function $f(x) + f^{-1}(x) \equiv 2x$ is strictly increasing. Since $f(x)$ is monotonic and invertible, $f(x)$ is strictly monotonic. Since $f(x)$ and $f^{-1}(x)$ are either both increasing or both decreasing, $f(x)$ is strictly increasing. Therefore, when $k$ is a non-zero integer, $f^{k}(x)$ is also strictly increasing. Finally, we prove that $g(x)$ is a constant function. Otherwise, there exist $x_{1}, x_{2} \in \mathbb{R}, x_{1} \neq x_{2}$ such that $g(x_{1})|x_{2} - x_{1}| > 0$,

i.e., $x_{1} - k g(x_{1}) > x_{2} - k g(x_{2})$,
or $x_{2} + k g(x_{2}) > x_{1} + k g(x_{1})$.
By equation (2), when $i=1,2$, we have
$$\begin{aligned}
f^{-k}(x_{i} - k g(x_{i})) & \equiv f^{-k+1}(x_{i} + (-k+1) g(x_{i})) \\
& \equiv f^{-k+2}(x_{i} + (-k+2) g(x_{i})) \\
& \equiv \cdots \equiv f^{-1}(x_{i} - g(x_{i})) \\
& \equiv x_{i}
\end{aligned}$$

Thus, we deduce that $x_{1} > x_{2}$, which contradicts $x_{2} - x_{1} > 0$. Similarly, we can prove that the latter case is also impossible. This proves that $g(x) \equiv c$, and thus
$$f(x) \equiv x + c, \quad x \in \mathbb{R}$$

Note that for any $c \in \mathbb{R}$, the function $f(x) \equiv x + c$ satisfies the conditions given in the problem.