Let $n$ be a natural number whose prime factorization is $$n=\prod_{i=1}^{k}p_i^{\alpha_i} \; .$$ Define a function $g(n)$ as follows $$g(n)=\sum_{i=1}^{k}p_i {\alpha_i} \,$$ i.e., exponentiation is "demoted" to multiplication, and multiplication is demoted to addition. For example: $n=20=2^2 5^1$, $f(n) = 2\cdot 2 + 5\cdot 1 = 9$.

Define $f(n)$ to repeat $g(n)$ until a cycle is reached. For example: $n=154=2^1 7^1 11^1$, $g(n)=20$, $g^2(n)=g(20)=9$, $g^3(n)=g(9)=6$, $g^4(n)=g(6)=5$, and now $g^k(n)=5$ for $k \ge 4$. So $f(154)=5$.

It is clear that every prime is a fixed point of $f(\;)$. I believe that $n=4$ is the only composite fixed point of $f(\;)$.

Q1. Is it the case that $4$ is the only composite fixed point of $f(\;)$, and that there are no cycles of length greater than $1$? (Yes: SeeEmilJeřábek's comment.)

Q2. Does every prime $p$ have an $n \neq p$ such that $f(n) = p$, i.e., is every prime "reached" by $f(\;)$? (Yes: SeeJeremyRouse's answer.)

There appear to be interesting patterns here. For example, it seems
that $f(n)=5$ is common.
(**Indeed**: See მამუკა ჯიბლაძე's graphical display.)