Any ideas how to prove that no injection $f : \mathbb{N} \rightarrow \mathbb{N}$ whose image is closed under multiplication by elements of $\mathbb{N}$ satisfies the following identity? $$f(n^m) = mf(n)$$
Its just a conjecture (I mean, it could be false).
Motivation. Let $A$ and $X$ denote sets. Suppose $\circ : A^2 \rightarrow A$ is associative, $f : X \rightarrow A$ is injective, and that the image of $f$ is left-closed under $\circ$. Then there exists a unique operation $* : A \times X \rightarrow X$ satisfying
$$f(a * x) = a \circ f(x).$$
Furthermore, it necessarily has the property that:
$$a * (b * x) = (a \circ b) * x.$$
I ask myself: does there exist a pair of operations $* : A \times X \rightarrow X$ and $\circ : A \times A \rightarrow A$ satisfying the above identity, such that no injection $f : X \rightarrow X$ whose image is left-closed under $\circ$ has the property that $f(a * x) = a \circ f(x)$?
I think the answer is "yes." Indeed, I think that taking $A,Y = \mathbb{N}$ and $$m * n = n^m, \;m \circ n = nm$$ should yield an example. Hence the question.
Suppose that $f$ is such a function.
$f(1) = f(1^2) = 2f(1)$ yields $f(1) = 0$. If you don't consider $0\in \mathbb{N}$ this is already a problem; otherwise $f(0) = f(0^2) = 2f(0)$ gives $f(0) = 0$ as well contradicting injectivity.