In the complex world there is a fundamental involution, namely the map $\gamma:\ z\mapsto\bar z$. An involution of some set $X$ is a map $\iota:\ X\to X$ which is not the identity map ${\rm id}_X$, but whose square $\iota\circ\iota$ is the identity.
Given any region $\Omega\subset{\mathbb C}$ and a function $f:\ \Omega\to{\mathbb C}$, one can compose $f$ with $\gamma$ in various ways. Your cases 2. and 3. produce for a given $f$ the functions
$$f\circ \gamma:\quad z\mapsto f\bigl(\bar z)$$
and
$$\gamma\circ f:\quad z\mapsto\overline{f(z)}\ .$$
When $f$ is a holomorphic function of $z$ then both $f\circ\gamma$ and $\gamma\circ f$ are antiholomorphic, which means, e.g., that nonoriented angles between tangent vectors at points $z_0\in\Omega$ are preserved, but the orientation of small circles around $z_0$ is reversed.
Most interesting is your case 1. Here the function $f$ is transformed (also termed conjugated) into the new function
$$\bar f:=\gamma\circ f\circ\gamma:\quad z\mapsto\overline{ f(\bar z)}\ .$$
When $f$ is a holomorphic function on $\Omega$ then it easy to check by means of the CR-equations and the chain rule that the function $\bar f$ is a holomorphic function on $\bar\Omega:=\{\bar z|z\in\Omega\}$.
The case when $\Omega$ and $\bar\Omega$ share an interval on the real axis is of special interest, because then we can ask the question: Could it be that in fact $\bar f(z)\equiv f(z)$ for all $z\in \Omega\cap\bar\Omega\ $? After all, in the definition of $\bar f$ there are two conjugations involved.
To investigate this case, assume that $0\in\Omega\cap\bar\Omega$ and that
$$f(z)=\sum_{k=0}^\infty a_kz^k\qquad(|z|<\rho)$$
with certain complex coefficients $a_k$. Then
$$\bar f(z)=\overline{\sum_{k=0}^\infty a_k(\bar z)^k}=\sum_{k=0}^\infty \bar a_k z^k\ ,$$
and this is $\ \equiv f(z)$ iff all $\bar a_k=a_k$, i.e., if all $a_k$ are in fact real. When the latter is the case then automatically $f(z)$ is real for real $z$, and it is not difficult to show the converse: If a holomorphic $f(z)$ is real for real $z$ (as in the case $f:=\sin$) then $\bar f=f$. This is the so-called reflection principle.
Best Answer
Let's approach it in the reverse order as Churchill. Suppose $f(z)$ is analytic within some domain $D$. Then, for any constant $c$, the function $g(z)=c^2/f(z)$ is analytic within the same domain as $f(z)$ (save at zeroes of $f(z)$, but we won't need to worry about that caveat.) So we have two functions which certainly satisfy the Cauchy-Riemann equations on $D$.
Note that nothing has been said about $\overline{f(z)}=\dfrac{|f(z)|^2}{f(z)}$. But suppose that the modulus of $f(z)$ is some nonzero constant within $D$ i.e. $|f(z)|=c\neq 0$ for some $c\in\mathbb{C}$, for all $z\in D$. Under that particular assumption, the $g(z)$ defined previously must be identical to $\overline{f(z)}$---and yet, by construction, it must also be analytic.
What this implies is that both $f(z)$ and its conjugate are analytic within $D$. But then the Cauchy-Riemann equations demand that $f(z)$ must be constant on this domain. (This result is simple to prove, and is presumably the Example 3 to which Churchill alludes.) Hence there are no non-constant functions which are both analytic and have constant modulus on a given domain.