Special bijections on $n$ elements

Question

Is there any bijection $f$ on $n$ elements s.t: $$f(1)=1$$ and
$$f(i)+j\neq f(j)+i\quad 1\leq i<j\leq n $$ ?

I found that it's always true for odd $n$, but it seems impossible when $n$ is an even number, but I can't prove this conjecture yet. What's your opinion?

EDIT: Indeed I actually want to a solution in $\mathbb I_n $(or$\mathbb Z/n\mathbb Z$) so I ask for a bijection s.t: $$f(1)=1$$ and
$$f(i)+j\neq f(j)+i\pmod n \quad 1\leq i<j\leq n $$
Like a circle!

Answer 1

Posting this as a separate answer since the OP became a very different question after the latest edit. In fact, the conjecture can now be proved to hold true, even without the assumption $f(1) = 1$.

Lemma 1. If $n$ is an even integer then $\sum_{k=1}^{n} k \equiv \frac{n}{2} \pmod n$.

This follows directly from the summation formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2} = \frac{n}{2} n + \frac{n}{2}\equiv \frac{n}{2} \pmod n$.

Lemma 2. If $g(k)$ is a permutation of $\{1, 2, ... n\}$ then $\sum_{k=1}^{n} g(k) \equiv \frac{n}{2} \pmod n$.

This follows directly from Lemma 1 since $\{g(k)\;|\;1 \le k \le n \} = \{1, 2, ... n\}$.

Now suppose that a permutation $f$ existed with the given properties. For the condition to hold $$g(k) = 1 + (f(k) - k \; \bmod n)$$ must be injective, therefore bijective. But $$\sum_{k=1}^{n} g(k) = n + \sum_{k=1}^{n}(f(k) - k) \; \bmod n = n$$

where the second term canceled out because $\sum_{k=1}^{n} f(k) = \sum_{k=1}^{n} k$ since $f$ itself is a permutation.

This leaves $\;\sum_{k=1}^{n} g(k) = n \not \equiv \frac{n}{2} \pmod n\;$ so by the contrapositive of Lemma 2 $g$ cannot be a bijection. Therefore a bijection $f$ does not exist with the given properties.

Answer 2

it seems impossible when $n$ is an even number, but I can't prove this conjecture

A counterexample for $n = 6$ is the permutation $ f = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 4 & 6 & 5 & 3 & 2 \\ \end{pmatrix} $.

$f(k) - k$ takes the distinct values $\{ 0, 2, 3, 1, -2, -4 \}$ so $f(k) - k$ is injective, therefore

$$ f(i) - i \ne f(j) - j \;\;\;\;\; 1 \le i \lt j \le 6 $$

which is equivalent to the given condition.