Consider an element $e \in \pi_n^s(S^0)$, in the stable homotopy groups of sphere. Let $C$ denote the spectrum which is cone of $e$, i.e. $C$ fits in the cofiber sequence
$$ S^n \to S^0 \to C.$$
$\textbf{Question:}$ I want to know under what condition $e$ viewed as a self map of $C^{\wedge k}$ (the $k$-fold smash product of C) is null-homotopic, in other words how do we detect if $e \wedge 1_{C^{\wedge k}}$ is homotopically trivial.
Let me expand on the case when $k=1$.
We have a cofiber sequence $$S^n \wedge C \to S^0 \wedge C \to C \wedge C $$ where the left most map is $e \wedge 1_C$. If $e \wedge 1_C$ is trivial then the $C$ splits of $C \wedge C$.
I believe that if $e \cup_1 e \neq 0$ then $e \wedge 1_C$ is non-trivial. The reason being (I think) $e \cup_1 e$ is the non-trivial attaching map for the top-cell of $C \wedge C/\Sigma_2$ to the $0$-cell. Let me know if I am stating something incorrect. I think the converse should be true as well.
$\textbf{Question:}$ If $e \cup_1 e \neq 0$ then can we say that $e \wedge 1_{C^{\wedge k}}$ is non-trivial for $k>1$ as well?
Further let me give an example, let $e = 2: S^0 \to S^0$. We know that $2$ is a non-trivial as a self-map of $M_2$ (the cone of $2$), the reason being $2 \cup_1 2 = \eta$.
$\textbf{Question:}$ Is $ 2 \wedge 1_{M_2^{\wedge k}}$ non-trivial for all $k$?
I believe the answer should be yes. Does anyone know of a proof?
$\textbf{Question:}$ Is there an example of an $e$ such that $e \wedge 1_{C^{\wedge k}}$ is non-trivial for $k <n$ but homotopically trivial for $k = n$, where n > 1?