If I have a scalar $x$ and point $B$, then I can compute $X = f(x,B)$
If the function $f$ is point multiplication, i.e. $f(x,B) = x \cdot B$, then $B$ can be determined if $X$ and $x$ are known.
Under the circumstances where $x$ and $X$ will be known, is it possible to modify the function $f$ such that $B$ cannot be determined?
It is necessary that:
the function $f$ is commutative, i.e. $f(x, f(y, B)) = f(y, f(x, B))$
$X$ cannot be determined from $x$.
If many pairs of ($x$, $X$) are given out, it cannot be inferred that any two of the pairs were created using the same base point $B$, even if $B$ is not itself determinable.
Edit: It would be fine to limit our choice of $x$ if that would prevent $B$ from being determined.
Edit: The curve used is ed25519