Subset of $\mathbb{Z}_k^n$ with sufficient local deviations is exponentially large?

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Given a non-empty set of vectors $S\subset\mathbb{Z}_k^n$. For every $x_1\in S$ and $i=1,\ldots,n$ there exist $x_2,\ldots,x_k\in S$ such that the vectors $x_1,\ldots,x_k$ are all the same in position $i$ and are all different in any other position. Let $f_k(n)$ denote the minimum possible size of $S$. Does $f_k(n)$ grow exponentially for all $k\geq2$?

For example if $k=2$ then $S$ is closed under adding any vector with all $1$s except for one $0$, the span of which has dimension $n$ for $n$ even and $n-1$ otherwise, giving $f_2(n)\geq2^{n-1}$, which is exponential. A similar argument gives $f_k(2)=k^2$ for all $k\geq2$.

However, for all $k\geq2$ it turns out that $f_k(n)$ is still only $k^2$ for all $2\leq n\leq k+1$. Indeed the span of $(0,1,\ldots,1)$ and $(1,0,1,2,\ldots,k-1)$ fulfills the given criteria.

But I still think $f_k(n)$ grows exponentially for all $k$. Maybe something like $f_k(n)\geq k^{n/k}$ still holds?