Let $R\ne\{0\}$ be a ring with identity and $2\leq n$ an integer.
For each $1\leq i\leq n$, let $\lambda_i:[1,n]\rightarrow R$ be the mapping for which $\lambda_i(i)=1$ and $\lambda_i(j)=0$ for $j\ne i$. Let $X_i:\mathbf{N}^n\rightarrow R$ be the mapping such that $X_i(\lambda_i)=1$ and $X_i(\alpha)=0$ for $\alpha\ne\lambda_i$.
Is it true that, for each $k\in\mathbf{N}$, we must have $$X_i(k\lambda_i)=kX_i(\lambda_i)$$ for all $1\leq i\leq n$? I think the answer is yes, but I am struggling to prove it by induction. It holds trivially for $k=0$. Let it hold for $k$. Then $$X_i((k+1)\lambda_i)=X_i(k\lambda_i+\lambda_i).$$ What to do next?
Let $k = 2$. \begin{align*} X_i(k \lambda_1) &= X_i(2 \lambda_i) = 0 \text{, and } \\ k X_i(\lambda_i) &= 2 \cdot 1 = 2 \text{.} \end{align*}
(The first is equal to $0$ because the various $X_i$ are only nonzero for certain functions whose range is $\{0,1\}$, but $2\lambda_i(i) = 2 \not\in \{0,1\}$.)