For some set $A$, I want to show that if there's a surjection from $A$ to the set $(A \Rightarrow A)$ of functions from $A$ to $A$, then $A$ has exactly 1 element.
Note that $(A \Rightarrow A)$ is the indicator/characteristic function. So we have $\mathcal{P}(A) = (A \Rightarrow [2])$ for example.
I think the difficulty is showing for all such functions $D$ has one element, so we need a proof by contradiction. If $D$ has no elements, then the state is false immediately since there cannot be surjection.
Then the interesting case lies when $D$ has at least 2 elements, where we need to find a contradiction. Since for all functions $D \to D$ must be bijective (I think) how can we proceed to find such a contradiction? I have run out of ideas from here.
The number of functions from $A \to A$ is $|A|^{|A|}$, the only way for $|A| \geq |A|^{|A|}$ is if $|A| = 1$.