$$\boxed{\binom{2022}{1011}}$$
We say that $x$ is a fixed point of $f$ if we have $f(x) = x$. Suppose that a function $f$ belonging to $\mathcal{A}$ has two fixed points $a$ and $b$ with $a < b$. Then, it follows from the first condition that $a = f(a) \ge f(b) = b$, which is a contradiction. Therefore, each function belonging to $\mathcal{A}$ has at most one fixed point.

Suppose that $\mathcal{A}$ is non-empty. Take a function $f$ belonging to $\mathcal{A}$, and define $a = f(f(1))$. Then, $a$ is a fixed point of $f$ since we have $f(f(1)) = f(f(f(1)))$ by the second condition. Let $g$ be a function belonging to $\mathcal{A}$. Then, we have
$$g(a) = g(f(a)) = f(g(f(a))) = f(g(a))$$
by the second condition. It shows that $g(a)$ is also a fixed point of $f$. By the uniqueness of the fixed point of $f$, we conclude that $g(a) = a$. Therefore, we have proved that $a$ is the unique fixed point of any function $g$ belonging to $\mathcal{A}$.
Let $f$ and $g$ be functions belonging to $\mathcal{A}$, and $x$ an integer from $1$ to $2023$. Then, the second condition says that $f(g(x))$ is a fixed point of $g$, and hence, we have $f(g(x)) = a$.
Let $X$ denote the set of integers $x$ from $1$ to $2023$ such that $f(x) = a$ holds for any function $f$ belonging to $\mathcal{A}$. Let $\ell$ be the smallest element of $X$, and $m$ the largest element of $X$. Then, for any function $f$ belonging to $\mathcal{A}$, we have $a = f(\ell) \ge f(\ell+1) \ge \dots \ge f(m) = a$ by the first condition. Therefore, we conclude that $X$ is the set of integers from $\ell$ to $m$. Note that we already proved that $f(g(x)) = a$ for any functions $f$ and $g$ belonging to $\mathcal{A}$ and any integer $x$ from $1$ to $2023$. Therefore, we conclude that $g(x)$ belongs to $X$ for such $g$ and $x$. Summarizing the discussion here, it has been shown that $f$ belonging to $\mathcal{A}$ satisfies both
$$(1) \quad m \ge f(1) \ge f(2) \ge \dots \ge f(2023) \ge \ell, \text{ and}$$
$$(2) \quad \text{we have } f(x) = a \text{ for any integer } x \text{ from } \ell \text{ to } m.$$
We fix a function $f$ belonging to $\mathcal{A}$. For an integer $x$ from $1$ to $2023$, we define $s_x = f(x) - x$. Then, by (1), the sequence $s_1, s_2, \dots, s_{2023}$ satisfies
$$m - 1 \ge s_1 > s_2 > \dots > s_{2023} \ge \ell - 2023.$$
Furthermore, by (2), we have $s_x = a - x$ for any integer $x$ from $\ell$ to $m$. Therefore, for any function that belongs to $\mathcal{A}$, we can associate it with a method of choosing $2023$ integers from the integers from $\ell - 2023$ to $m - 1$, such that all integers from $a - m$ to $a - \ell$ are chosen. Here, this choice corresponds to choosing $2023 - ((a - \ell) - (a - m) + 1) = 2022 - m + \ell$ integers from the $2022$ integers that are from $\ell - 2023$ to $m - 1$ but not between $a - m$ and $a - \ell$. Hence, the number of such choice is $\binom{2022}{2022 - m + \ell}$. Furthermore, since the methods of choosing corresponding to different functions belonging to $\mathcal{A}$ are distinct, the number of elements in $\mathcal{A}$ is at most $\binom{2022}{2022 - m + \ell}$. In addition, for a positive integer $x$ from $0$ to $2021$, we have
$$\frac{\binom{2022}{\ell+1}}{\binom{2022}{\ell}} = \frac{2022 - \ell}{\ell+1},$$
which implies
$$\binom{2022}{0} < \binom{2022}{1} < \dots < \binom{2022}{1010} < \binom{2022}{1011} > \binom{2022}{1012} > \dots > \binom{2022}{2021} > \binom{2022}{2022}.$$
Therefore, we can conclude that the number of elements in $\mathcal{A}$ is at most $\binom{2022}{1011}$.
On the other hand, consider the set $\mathcal{B}$ of functions $f$ that are defined for integers from $1$ to $2023$ and take integer values from $1$ to $2023$, and satisfy
$$1012 = f(1) = f(2) = \dots = f(1012) \ge f(1013) \ge f(1014) \ge \dots \ge f(2023).$$
This set satisfies the first condition. Furthermore, for any functions $f$ and $g$ in $\mathcal{B}$ and any integer $x$ from $1$ to $2023$, we have $g(x) \le 1012$, so $f(g(x)) = 1012$ holds. Thus, the second condition is also satisfied. By considering $\ell = 1$ and $a = m = 1012$ in the argument above, we see that there is a one-to-one correspondence between the functions $f$ in $\mathcal{B}$ and the sequences $s_1, s_2, \dots, s_{2023}$ of integers satisfying
$$s_1 = 1011, s_2 = 1010, \dots, s_{1012} = 0, \quad -1 \ge s_{1013} > s_{1014} > \dots > s_{2023} \ge -2022.$$
Therefore, the number of elements in $\mathcal{B}$ is $\binom{2022}{1011}$.
Therefore, we conclude that the maximum possible number of elements in $\mathcal{A}$ is $\binom{2022}{1011}$.