Existence of a meromorphic functions $f(z)$ such that $|f(z)|\geq |z|$.

Let $f$ be a meromorphic function on $\mathbb{C}$ such that $|f(z)|\geq |z|$ at each $z$, where $f$ is holomorphic. Then, which of the following is/are true.

1. The hypothesis are contradictory, so no such $f$ exists.
2. Such an $f$ is entire.
3. There is a unique $f$ satisfying the given conditions.
4. There is an $A\in \mathbb{C}$ with $|A|\geq 1$ such that $f(z)=Az$ for each $z\in \mathbb{C}$

I think $1$ and $3$ are false but not sure. And what about $2$ and $4$? I am unable to find suitable examples. Please help me here.

It should be part of the problem statement that $f$ is meromorphic on all of $\mathbb{C}$, we could not say much about $f$ if we only assume it is meromorphic on $\{z : \lvert z-5\rvert < 4\}$ for example.

Let's look at the slightly more general situation that we are given two functions $g,h$ that are meromorphic on all of $\mathbb{C}$. Define $P(g)$ as the set of poles of $g$ and $P(h)$ analogously. Assume further that we have an inequality

$$\lvert g(z)\rvert \leqslant \lvert h(z)\rvert \tag{1}$$

for all $z$ such that neither $g$ nor $h$ has a pole at $z$.

If we know that $h \not\equiv 0$ - which is for example implied by $g\not\equiv 0$ - then we can look at the quotient

$$q(z) = \frac{g(z)}{h(z)},$$

which we know is defined and holomorphic at least on $\mathbb{C}\setminus E$, where $E = P(g) \cup P(h) \cup h^{-1}(0)$. Since $h$ is not identically zero, $h^{-1}(0)$ is a closed and discrete subset of $\mathbb{C}$. As a union of finitely many closed and discrete sets, $E$ is a closed and discrete subset of $\mathbb{C}$, so $q$ is holomorphic except for isolated singularities on all of $\mathbb{C}$. From $(1)$, we immediately obtain $\lvert q(z)\rvert \leqslant 1$ on $\mathbb{C}\setminus E$, so by Riemann's removable singularity theorem, all the isolated singularities of $q$ are removable, and $q$ has an extension to an entire function (which we again denote by $q$). The inequality $\lvert q(z)\rvert \leqslant 1$ extends from $\mathbb{C}\setminus E$ to $\mathbb{C}$ by continuity, and hence Liouville's theorem tells us that $q$ is constant, $q(z) \equiv c$, where $\lvert c\rvert \leqslant 1$.

Thus $(1)$ implies that $g \equiv c\cdot h$ for some $c$ with $\lvert c\rvert \leqslant 1$. The argument above requires $h \not \equiv 0$, but the case $h \equiv 0$ immediately implies $g \equiv 0$, i.e. $g \equiv 1\cdot h$ (or $g \equiv 0\cdot h$, in this case $c$ isn't unique).

Applying this to the given situation with $g(z) = z$ should make it clear which of the points 1.-4. are true and which are false.