There are two solutions:
* $f(a, b) = a$ for all $a, b$, and
* $f(a, b) = b$ for all $a, b$.
Clearly these functions meet the condition. We must show there are no others.

By setting $a = b = c$ we get $f(a, a) = a$ for all $a$. Next, for all $a, b$, the median of $f(a, a)$, $f(a, b)$, $f(b, a)$ must equal the median of $a, a, b$, namely $a$, so for all $a, b$, one of $f(a, b)$, $f(b, a)$ is at most $a$ and the other is at least $a$. Switching $a, b$, we also see that one of $f(a, b)$, $f(b, a)$ is at most $b$ and the other is at least $b$. Therefore, we have
$$\min\{f(a, b), f(b, a)\} \le \min\{a, b\} \qquad (13)$$
and
$$\max\{f(a, b), f(b, a)\} \ge \max\{a, b\}. \qquad (14)$$
Next, consider any three numbers $a < b < c$. The median of $f(a, b)$, $f(b, c)$, $f(c, a)$ must equal $b$, so one of $f(a, b)$, $f(b, c)$ equals $b$ (since $f(c, a)$ must be either at most $a$ or at least $c$). Similarly, considering $f(a, c)$, $f(c, b)$, $f(b, a)$, we see that one of $f(c, b)$, $f(b, a)$ equals $b$. The numbers $f(a, b)$, $f(b, a)$ cannot both be $b$, by (13), and $f(b, c)$, $f(c, b)$ cannot both be $b$, by (13). We conclude that either
$$f(a, b) = f(c, b) = b \qquad (15)$$
or
$$f(b, c) = f(b, a) = b. \qquad (16)$$
In particular, for any $a < b$, choosing $c > b$ arbitrarily, we see that one of $f(a, b)$, $f(b, a)$ must equal $a$. Likewise, for any $b < c$, choosing $a < b$ arbitrarily, we see that one of $f(b, c)$, $f(c, b)$ must equal $b$.

Putting these two conclusions together, for any $a \ne b$, one of $f(a, b)$, $f(b, a)$ equals $\min\{a, b\}$ and the other equals $\max\{a, b\}$. In other words, for $a \ne b$, $\{f(a, b), f(b, a)\}$ and $\{a, b\}$ are equal as sets. Call $\{a, b\}$ a first-pair if $f(a, b) = a$ and $f(b, a) = b$, and a second-pair if $f(a, b) = b$ and $f(b, a) = a$.

Now again consider any three numbers $a < b < c$. If either $\{a, b\}$ or $\{b, c\}$ is a first-pair, then (15) cannot hold, so (16) holds, and $\{a, b\}$ and $\{b, c\}$ are both first-pairs. That is, $\{a, b\}$ is a first-pair if and only if $\{b, c\}$ is. Pick any other numbers $a'$ and $c'$ such that $a' < b$ and $c' > b$. The same logic gives
$$\begin{align*} \{a', b\} \text{ is a first-pair} &\iff \{b, c\} \text{ is a first-pair} \\ &\iff \{a, b\} \text{ is a first-pair} \\ &\iff \{b, c'\} \text{ is a first-pair.} \end{align*}$$
This shows that, given any $p, q \ne b$, $\{p, b\}$ is a first-pair if and only if $\{q, b\}$ is. Since $b$ is arbitrary, we have for any distinct $p, q, r, s$ that
$$\begin{align*} \{p, q\} \text{ is a first-pair} &\iff \{s, q\} = \{q, s\} \text{ is a first-pair} \\ &\iff \{r, s\} \text{ is a first-pair.} \end{align*}$$
That is, if some first-pair exists, then every pair is a first-pair, so $f(a, b) = a$ for all $a, b$. Otherwise, every pair is a second-pair, so $f(a, b) = b$ for all $a, b$. Thus, the only possibilities for the function $f$ are the two solutions we initially identified.