Linear constraint equivalent to one-to-oneness

Question

Suppose that $A$ is a finite set with cardinality $n$. Let $\phi$ be a one-to-one function from $A$ to $\{1,2,\dots,n\}$. I am wondering if it is possible to write the one-to-oneness condition on $\phi$ to an equivalent linear equality (or inequality) constraint on $\phi$. Any help is greatly appreciated.

Answer 1

The answer to your question is NO. In fact it already fails for $n=2$. In this case, $A=\lbrace a_1,a_2 \rbrace $ and there are exactly two one-to-one functions $A \to \lbrace 1,2 \rbrace $, namely $\phi_1$ defined by $\phi_1(a_1)=1,\phi_1(a_2)=2$ and $\phi_2$ defined by $\phi_2(a_1)=2,\phi_2(a_2)=1$.

Now, suppose a set of linear constraints is satisfied by both $\phi_1$ and $\phi_2$. Because the constraints are linear, they will still be satisfied by any linear combination of $\phi_1$ and $\phi_2$ with nonnegative coefficients. In particular, they will be satisfied by the mean $\frac{\phi_1+\phi_2}{2}$ which is constant equal to $1$ and therefore not one-to-one.

Similarly, for arbitrary $n$ there are $n!$ different one-to-one functions $A \to \lbrace 1,2,\ldots,n \rbrace $ and their (scaled) mean, i.e. their sum divided by $n$, is constant equal to $1$.

Answer 2

A bijection from A to S = { 1,2,.. n } usually it isn't linear.
How could it be if A, for example, were a set of open sets?
Even if A were a set of integers,
how could there be a linear bijection from { x$^2$ : x in S } to S?