The empty theory does not uniformly eliminate imaginaries. The empty theory $\varnothing$ is the theory of a single set. We have $=$, but only because its a logical symbol of FOL. I'm picking this theory because it's the simplest theory there is.
I'll take the following as the definition of uniform elimination of imaginaries, from page 117 of A shorter model theory. I asked about this definition in an earlier question here.

Another way of saying this [that $A$ has uniform elimination of imaginaries] is that for every equivalence formula $\theta(\vec{x}, \vec{y})$ of $A$ there is a function $F$ which is definable without parameters, taking tuples as values, such that for all $\vec{a_1}$ $\vec{a_2}$, $\vec{a_1}$ is $\theta$-equivalent to $\vec{a_2}$ if and only if $F(\vec{a_1}) = F(\vec{a_2})$.

My question is: is the following argument correct for showing that the empty theory does not have uniform elimination of imaginaries?
Theorem: The empty theory does not uniformly eliminate imaginaries.
Proof.
Let $L$ be the empty language.
Let $A$ be the $L$-structure over the domain $\{0, 1\}$.
The relation $R$, defined as $(x, y \mapsto \top)$, is definable.
The language $L$ has no function symbols, so there are thus exactly two possibilities for $F$.

*

*$F_1$ is the identity function. $F_1(x) = y \iff x = y$

*$F_2$ swaps $0$ and $1$. $F_2(x) = y \iff x \neq y$
Neither $F_1$ nor $F_2$ are constant functions, therefore $x, y \mapsto F_\square(x) = F_\square(y)$ is never $R$.
Therefore $A$ does not uniformly imaginaries.
A theory uniformly eliminates imaginaries if and only if all of its models uniformly eliminate imaginaries.
Therefore, the empty theory does not uniformly eliminate imaginaries.
 A: Your argument is incorrect, on a technicality. Remember that $F$ should be a definable function $A\to A^n$ for some $n\in \mathbb{N}$, and this includes the case $n = 0$. Note that $A^0$ is a singleton set, and the formula $\top$ defines a (constant) function $A\to A^0$.
Hodges explicitly includes this trivial case in his definition: The paragraph defining elimination of imaginaries ends "We allow $\overline{z}$ and $\overline{b}$ to be empty here, in which case $\overline{a}/\theta$ is $\varnothing$-definable."
As a consequence, every theory uniformly eliminates imaginaries for the trivial equivalence relation $\top$. Every theory also uniformly eliminates imaginaries for the discrete equivalence relation $x = y$, by the identity function $F(x) = x$.
The next simplest definable equivalence relation is the "ordered pair" relation, defined by: $$(x,y)E_2(x',y') \iff ((x = x')\land (y = y'))\lor ((x = y')\land (y = x')).$$
Many theories, including the empty theory, fail to uniformly eliminate imaginaries for this equivalence relation.
To see this in the case of the empty theory, consider a set $A$ with at least $4$ elements, and suppose for contradiction that $F$ is a $\varnothing$-definable function such that $F(a,b) = F(c,d)$ if and only if $\{a,b\} = \{c,d\}$. The outputs of $F$ cannot be in $A^0$, since $F$ is not a constant function ($A/E_2$ has more than one class). Fix $a,b\in A$. If the tuple $F(a,b)$ contains any $c\notin \{a,b\}$, we can find an automorphism of $A$, call it $\sigma$, fixing $a$ and $b$ and moving $c$. Then $$F(a,b) = F(\sigma(a),\sigma(b)) = \sigma(F(a,b))\neq F(a,b),$$ contradiction. So $F(a,b)$ is a non-empty tuple such that every component is either $a$ or $b$.  Now let $\tau$ be the automorphism of $A$ swapping $a$ and $b$. We have $$F(b,a) = F(\tau(a),\tau(b)) = \tau(F(a,b)) \neq F(a,b),$$ contradiction.

It's worth mentioning that the empty theory does have a property called "weak elimination of imaginaries". We say that $T$ weakly eliminates imaginaries if every equivalence class for a definable equivalence relation is definable, but rather than requiring the parameter defining the class to be unique, we require that there are finitely many choices of parameters which work.
Weak elimination of imaginaries essentially means "elimination of imaginaries except maybe for the equivalence relations $(x_1,\dots,x_n)E_n(y_1,\dots,y_n)\iff \{x_1,\dots,x_n\} = \{y_1,\dots,y_n\}$. It's a theorem that if $T$ has weak elimination of imaginaries and also eliminates the imaginaries for the equivalence relations $E_n$ defined in the previous paragraph, the it has full elimination of imaginaries.
