Define a sequence $(p_i)_{i \ge 0}$ by letting $p_0 = p$ and $p_i = f^{(i)}(p)$ for $i \ge 1$. Putting $x = p_n$ in the given equation, $f^{(2)}(x) = s f(x) + t x$, gives the following linear recurrence
$$p_{n+2} = s p_{n+1} + t p_n. \qquad (25)$$
Applying $f^{(i)}$ for $i \ge 1$ to the condition $f^{(n+m)}(p) = f^{(m)}(p)$ implies $p_{n+k} = p_k$ for all $k \ge m$, i.e. the sequence $(p_i)_{i \ge 0}$ is periodic. We wish to prove $p_0 = p_1$. The characteristic polynomial of the linear recurrence (25) is $X^2 - sX - t$, with discriminant $s^2 + 4t$.

Case 1: $s^2 + 4t = 0$. In this case, the characteristic polynomial has a double root, say $r_0$. Since $r_0^2 = -t \ne 0$, we have $r_0 \ne 0$. The sequence then has the form $p_k = (\alpha + k\beta) r_0^k$ for $k \ge 0$. The condition $p_{n+k} = p_k$ for all $k \ge m$ then translates into
$$(\alpha + (n + k)\beta) r_0^{n+k} = (\alpha + k\beta) r_0^k \quad \text{for } k \ge m,$$
which could be rewritten as $k\beta(r_0^n - 1) = \alpha - (\alpha + n\beta) r_0^n$. Since this holds true for at least two positive values of $k$, we must have $\beta(r_0^n - 1) = 0$, i.e. $\beta = 0$ or $r_0^n = 1$.
If $r_0^n = 1$, the previous equation implies $n\beta = 0$, hence $\beta = 0$. If $\beta = 0$, the previous equation implies $\alpha(r_0^n - 1) = 0$. If $\alpha = 0$, we obtain $p_i = 0$ for all $i \ge 0$, in particular $p_0 = p_1$. Otherwise, we need to have $r_0^n = 1$. We can conclude that, unless $\alpha = \beta = 0$, we need to have $\beta = 0$ and $r_0^n = 1$. This gives $p_k = \alpha r_0^k$ for all $k \ge 0$, in particular $p_0 = \alpha \ge 0$ and $p_1 = \alpha r_0 \ge 0$. Therefore $r_0$ is a non-negative real number (we continue to assume $\alpha \ne 0$). Since $r_0^n = 1$ this implies $r_0 = 1$, and so $p_k = \alpha$ for all $k \ge 0$.

Case 2: $s^2 + 4t \ne 0$. In this case, the characteristic polynomial has two distinct roots $r_1 \ne r_2$. They are non-zero, because $r_1 r_2 = -t \ne 0$. The sequence then has the form $p_k = \alpha r_1^k + \beta r_2^k$ for $k \ge 0$, and the condition $p_{n+k} = p_k$ for all $k \ge m$ translates into $\alpha r_1^{n+k} + \beta r_2^{n+k} = \alpha r_1^k + \beta r_2^k$ for $k \ge m$. This can be rewritten as
$$r_1^k \alpha (r_1^n - 1) + r_2^k \beta (r_2^n - 1) = 0. \quad (26)$$
If $\alpha = \beta = 0$, we clearly have $p_0 = p_1 = 0$. From now on, we assume $(\alpha, \beta) \ne (0, 0)$. There are three cases: $r_1^n \ne 1$, $r_2^n \ne 1$, and $r_1^n = r_2^n = 1$. However, symmetry between $r_1$ and $r_2$ covers the second case once we prove the first one.

Case 2A: $r_1^n \ne 1$. We can rewrite (26) as
$$\alpha \left(\frac{r_1}{r_2}\right)^k = -\frac{\beta(r_2^n - 1)}{r_1^n - 1} \quad \text{for } k \ge m$$
and obtain $\alpha = 0$. Otherwise the non-zero LHS would give different values for different $k \ge m$, because $r_1 \ne r_2$ in Case 2. With $\alpha = 0$, we get $p_0 = \beta > 0$ and $p_1 = \beta r_2 \ge 0$. In particular, $r_2 \ge 0$ is a real number; in fact, $r_2 > 0$ since $r_2$ is non-zero. On the other hand, (26) with $\alpha = 0$ and $\beta \ne 0$ implies $r_2^n = 1$, hence $r_2 = 1$ and $p_0 = p_1$, as required.

Case 2B: $r_1^n = r_2^n = 1$. In this case, $r_1$ and $r_2$ are complex numbers that satisfy $|r_1| = |r_2| = 1$. We recall that $r_1 r_2 = -t$, hence we have $(-t)^n = r_1^n r_2^n = 1$. Since $t$ is a real number, we need to have $t = \pm 1$. If $t = 1$, then $r_2 = -1/r_1 = -\bar{r}_1$, which implies $s = r_1 + r_2 = r_1 - \bar{r}_1 = 2\Im(r_1)i$. As $s$ is real, this is only possible if the imaginary part of $r_1$ is equal to zero, but then $s = 0$, which was excluded. Hence, we must have $t = -1$, which implies $r_2 = 1/r_1 = \bar{r}_1$.
If we let $r_1 = u + iv$ with real numbers $u, v$, then $r_2 = u - iv$ and
$$p_0 = \alpha + \beta \\ p_1 = \alpha r_1 + \beta r_2 = (\alpha + \beta)u + (\alpha - \beta)vi.$$
Since $p_1$ is real, we need to have $(\alpha - \beta)v = 0$. If $v = 0$, then $r_1 = r_2$, which was excluded in Case 2. Hence $\alpha = \beta$ and
$$p_k = \alpha(r_1^k + r_2^k) = \alpha(r_1^k + \bar{r}_1^k) = 2\alpha\Re(r_1^k).$$
Because $r_1$ is an $n$-th root of unity, there exists a smallest positive integer $d$ for which $r_1^d = 1$, and $r_1$ is a primitive $d$-th root of unity. As $r_1 \neq r_2$ we have $d > 1$. A complete list of $d$-th roots of unity then is
$$r_1, r_1^2, r_1^3, \dots, r_1^{d-1}, r^d.$$
Since $p_0 = \alpha > 0$ and $p_k \ge 0$ for all $k \ge 0$, the real parts of all these $d$-th roots of unity need to be non-negative. However, for each $d \ge 2$ there is a $d$-th root of unity with negative real part. This can be seen explicitly by considering the $d$-th roots of unity in trigonometric form
$$\zeta_k = \cos \left( \frac{2k\pi}{d} \right) + i \sin \left( \frac{2k\pi}{d} \right) \quad k = 0, 1, \dots, d-1.$$
The real part of $\zeta_k$ is equal to $\cos(\frac{2k\pi}{d})$ and this is negative for $k = \lfloor \frac{d}{2} \rfloor$ if $d \ge 2$. We can conclude now that Case 2B is not possible, and we have shown in all other cases that $p_0 = p_1$, i.e. $f(p) = p$.

Since $x = (f^{(2)}(x) - s f(x))/t$, it is clear that $f$ is injective. This allows us to “peel off” layers of $f$ from the assumed equation
$$f^{(n+m)}(p) = f^{(m)}(p)$$
to deduce that
$$f^{(n)}(p) = p. \qquad (27)$$
Thus, we need to show that every periodic point is fixed. We assume that $n \in \mathbb{N}$ is minimal in (27).
From now on, a subscripted variable will denote an iterate, so $x_k = f^{(k)}(x)$, $p_k = f^{(k)}(p)$, etc., for every $k \ge 0$; in particular, $x_0 = x$ and $x_1 = f(x)$.
Inductively, we see that $p_{n+m} = p_m$ for all $m \in \mathbb{N}$. Putting $x = p_m$ for each $0 \le m < n$ in $f^{(2)}(x) = s f(x) + t x$, and adding these equations, we get
$$\sigma = (s + t) \sigma$$
where $\sigma := \sum_{i=0}^{n-1} p_i$. We conclude that either $\sigma = 0$ or $s + t = 1$.
By minimality of $n$, the points $p_i$, $0 \le i < n$, are distinct, so if $\sigma = 0$, then $n = 1$ and $p = 0$. This proves the desired result except in the case $s + t = 1$.

From now on, we assume $s + t = 1$. For the sake of contradiction assume that the least period $n$ of $p$ strictly exceeds 1. We rewrite $f^{(2)}(x) = s f(x) + t x$ as
$$f^{(2)}(x) = (1 - t) f(x) + t x, \quad x \in \mathbb{R}. \qquad (28)$$
This equation allows us to write all higher iterates in a similar form:
$$x_{k+1} = s_k x_1 + t_k x_0, \quad x \in \mathbb{R}, k \in \mathbb{N}. \qquad (29)$$
Now,
$$x_{k+2} = s_k x_2 + t_k x_1 = (s_k(1 - t) + t_k) x_1 + s_k t x_0.$$
so we get the recurrence relations
$$\left. \begin{array}{l} s_{k+1} = s_k(1 - t) + t_k \\ t_{k+1} = s_k t \end{array} \right\} \qquad (30)$$
Together with $s_1 = 1 - t$, $t_1 = t$, (30) defines $s_k$ and $t_k$ for all $k \in \mathbb{N}$. Note that $s_{k+1} + t_{k+1} = s_k + t_k$, so we have $s_k + t_k = 1$ for all $k \in \mathbb{N}$. Let us also use (30) to write $t_{k+2}$ in terms of $s_k$ and $t_k$. Since $t_{k+2} = s_{k+1} t$, we obtain
$$t_{k+2} = s_k t (1 - t) + t_k t \qquad (31)$$
Applying the identity $x_k = (1 - t_k) x_1 + t_k x_0$ with $x = p$ and $k = n$, we deduce
$$(1 - t_n) p_0 = (1 - t_n) p_1,$$
giving a contradiction (since the least period of $p$ strictly exceeds 1) unless $t_n = 1$. To finish the proof, we prove that $t_k$ never equals 1.
We assume without loss of generality that $k > 1$. We are given that $s, t \neq 0$. Since now $s = 1 - t$, we have $t \neq 0, 1$. These omitted values break the line into three intervals and we consider each separately.
Suppose first that $0 < t < 1$. In (30), $s_{k+1}$ and $t_{k+1}$ are linear combinations of $s_k$ and $t_k$, where the coefficients are positive. Since $s_1, t_1 > 0$, it follows inductively that $s_k, t_k > 0$ for all $k \in \mathbb{N}$. Since $s_k + t_k = 1$, we deduce that $t_k \neq 1$.
Suppose next that $t < 0$, and so $s_1 = 1 - t > 1$. Using (30), it follows by induction that $t_k < 0$ and $s_k > 1$ for all $k \in \mathbb{N}$.
Finally, we consider $t > 1$, and so $s_1 = 1 - t < 0$. By (30), we have $t_2 < 0$ and so $s_2 > 1$. It now follows inductively from (31) that $t_{2j} < 0$ and $s_{2j} > 1$ for all $j \in \mathbb{N}$. We deduce from (30) that $t_{2j+1} > 1$, and so $s_{2j+1} < 0$, for all $j \in \mathbb{N}$. This finishes the proof that $t_k$ never equals 1, and so we are done.