Example 7 Given $f(f(f(x)))=8x+7$, find the linear function $f(x)$ with real coefficients. (1979 Anhui Competition Question)
Let $f(x) = ax + b (a \neq 1, a, b \in \mathbf{R})$ be a linear function with real coefficients. Then, $a^{3}\left(x - \frac{b}{1-a}\right) + \frac{b}{1-a} = 8x + 7$,
which implies $\left\{\begin{array}{l}a^{3} = 8 \\ \frac{(a^{3} - 1)b}{1-a} = -7\end{array} \Rightarrow \left\{\begin{array}{l}a = 2, \\ b = 1.\end{array}\right.\right.$
Therefore, the required linear function with real coefficients is $f(x) = 2x + 1$.
In Theorem 1, let $x_{0} = \frac{b}{1-a}$, then $f^{(n)}(x) = a^{n}(x - x_{0}) + x_{0} = \varphi^{-1} \circ g^{(n)} \circ \varphi(x)$. (where $\varphi(x) = x - x_{0}, g(x) = ax, g^{(n)}(x) = a^{n}x$, and “.” denotes the function composition operation.)

Thus, for the given functions $f(x)$ and $g(x)$, if there exists an invertible function $\varphi(x)$ such that $f(x) = \varphi^{-1} \circ g \circ \varphi(x)$, then $f(x)$ and $g(x)$ are said to be similar with respect to the bridge function $\varphi(x)$, denoted as $f \sim g$.
In this case, from $f(x) = \varphi^{-1}(g(\varphi(x)))$ we have $\varphi(f(x)) = g(\varphi(x))$, so $\varphi(x_{0}) = \varphi(f(x_{0})) = g(\varphi(x_{0}))$, which means $\varphi(x_{0})$ is a fixed point of $g(x)$, indicating that the bridge function $\varphi$ has the property: it maps the fixed point $x_{0}$ of $f$ to the fixed point $\varphi(x_{0})$ of $g$.

Typically, to facilitate the calculation of $g^{(n)}(x)$, $g(x)$ is often chosen as $ax, x + a, ax^2, ax^3$, etc., in which case the fixed points of $g(x)$ are 0 or $\infty$. Therefore, under the mapping of the bridge function $\varphi$, it is best to map the fixed point of $f$ to 0 or $\infty$. Thus, if $f(x)$ has a unique fixed point $x_{0}$, then we can consider taking $\varphi(x) = x - x_{0}$ (or $\frac{1}{x - x_{0}}$), in which case $\varphi(x_{0}) = 0$ (or $\infty$); if $f(x)$ has two distinct fixed points $x_{1}, x_{2}$, then we can consider taking $\varphi(x) = \frac{x - x_{1}}{x - x_{2}}$, in which case $\varphi(x_{1}) = 0, \varphi(x_{2}) = \infty$. Therefore, Theorem 1 can be strengthened and stated as follows:

Theorem 2 For a linear function $f(x) = ax + b (a \neq 1)$, if we take $\varphi(x) = x - x_{0}$ and $g(x) = ax$, then $f \stackrel{\varphi}{\sim} g \Leftrightarrow x_{0}$ is a fixed point of $f(x)$, and $f^{(n)}(x) = \varphi^{-1}(g^{(n)}(\varphi(x)))$.
Similarly, we can derive a series of theorems and corollaries:
Theorem 3 For a quadratic function $f(x) = ax^2 + bx + c (a \neq 0)$, if we take $g(x) = ax^2$ and $\varphi(x) = x - x_{0}$, then $f \stackrel{\varphi}{\sim} g \Leftrightarrow x_{0} = -\frac{b}{2a}$ and $x_{0}$ is a fixed point of $f(x)$, and $f^{(n)}(x) = \varphi^{-1}(g^{(n)}(\varphi(x)))$.
Theorem 4 For a linear fractional function $f(x) = \frac{ax + b}{cx + d} (c \neq 0, ad - bc \neq 0)$.
(1) If we take $\varphi(x) = \frac{x - x_{1}}{x - x_{2}} (x_{1} \neq x_{2})$ and $g(x) = kx$, then $f \sim g \Leftrightarrow x_{1}, x_{2}$ are two distinct fixed points of $f(x)$, and $k = \frac{a - cx_{1}}{a - cx_{2}}$, and $f^{(n)}(x) = \varphi^{-1}(g^{(n)}(\varphi(x)))$.
(2) If we take $\varphi(x) = \frac{1}{x - x_{0}}$ and $g(x) = x + k$, then $f \stackrel{\varphi}{\sim} g \Leftrightarrow x_{0}$ is the unique fixed point of $f(x)$ and $k = \frac{2c}{a + d}$, and $f^{(n)}(x) = \varphi^{-1}(g^{(n)}(\varphi(x)))$.

Theorem 5 For a quadratic fractional function $f(x) = \frac{ax^2 + bx + c}{dx^2 + ex + f} (d$ and $ae$ are not both zero), if we take $\varphi(x) = \frac{x - x_{1}}{x - x_{2}} (x_{1} \neq x_{2})$ and $g(x) = kx^2$, then $f \stackrel{\varphi}{\sim} g \Leftrightarrow k = \frac{a - dx_{1}}{a - dx_{2}}, be = 4dc, e(e - 2a) = 2d(2f - b), x_{1}, x_{2}$ are two distinct fixed points of $f(x)$, and when $d \neq 0$, the third fixed point of $f(x)$ is $-\frac{e}{2d}$, and $f^{(n)}(x) = \varphi^{-1}(g^{(n)}(\varphi(x)))$.

Corollary 1 For $f(x) = \frac{ax^2 + bx + c}{ex + f} (ae \neq 0)$, if we take $\varphi(x) = \frac{x - x_{1}}{x - x_{2}} (x_{1} \neq x_{2})$ and $g(x) = x^2$, then $f \stackrel{\varphi}{\sim} g \Leftrightarrow b = 0, e = 2a$ and $x_{1}, x_{2}$ are two distinct fixed points of $f(x)$.

Corollary 2 For $f(x) = \frac{ax^2 + bx}{dx^2 + ex + f} (bd \neq 0)$, if we take $\varphi(x) = \frac{x - x_{1}}{x - x_{2}} (x_{1} \neq x_{2})$ and $g(x) = kx^2$, then $f \stackrel{\varphi}{\sim} g \Leftrightarrow k = \frac{a - dx_{1}}{a - dx_{2}}, e = 0, b = 2f$ and $x_{1}, x_{2}$ are two distinct non-zero fixed points of $f(x)$.

Corollary 3 For $f(x) = \frac{ax^2}{dx^2 + ex + f} (adf \neq 0)$, if we take $\varphi(x) = \frac{x - x_{0}}{x} (x_{0} \neq 0)$, fixed point.

Theorem 6 For $f(x) = \frac{ax^3 + bx^2 + cx + d}{ex^2 + fx + g} (ae \neq 0)$, if we take $\varphi(x) = \frac{x - x_{1}}{x - x_{2}} (x_{1} \neq x_{2})$ and $g(x) = x^3$, then $f \stackrel{\varphi}{\sim} g \Leftrightarrow b = 0, e = 3a, cf = 9ad, f^2 = 3a(3g - c)$, and $x_{1}, x_{2} (x_{1} \neq x_{2})$ and $-\frac{f}{6a}$ are three fixed points of $f(x)$, and $f^{(n)}(x) = \varphi^{-1}(g^{(n)}(\varphi(x)))$.

Corollary 4 For $f(x) = \frac{ax^3 + bx^2 + cx}{ex^2 + fx + g} (aec \neq 0)$, if we take $\varphi(x) = \frac{x - x_{1}}{x - x_{2}} (x_{1} \neq x_{2})$ and $g(x) = x^3$, then $f \sim \underset{\sim}{\sim} g \Leftrightarrow b = f = 0, e = 3a, c = 3g$ and $x_{1}, x_{2}$ are two distinct non-zero fixed points of $f(x)$.
Clearly, Theorems 4 to 6 are special cases of the following Theorem 7.
Theorem 7 For $f(x) = \frac{a(x - b)^k - b p(x - a)^k}{(x - b)^k - p(x - a)^k} (a \neq b, k \in \mathbf{N})$, if we take $\varphi(x) = \frac{x - a}{x - b}$ and $g(x) = px^k$, then $f \stackrel{\varphi}{\sim} g \Leftrightarrow a, b$ are two distinct fixed points of $f(x)$ and $f^{(n)}(x) = \varphi^{-1}(g^{(n)}(\varphi(x)))$.