If a transformation of a random variable preserves its mean and variance, can it change the covariance with another random variable?

Question

Let $X$ and $Y$ be two real-valued random variables in $L^2$, and let $f:\mathbb R\to\mathbb R$ be a deterministic function such that $\mathbb E[f(X)] = \mathbb E[X]$ and $\mathrm{Var}(f(X)) = \mathrm{Var}(X)$.

Does $$\mathrm{Cov}(f(X),Y) = \mathrm{Cov}(X,Y)$$ hold necessarily?

Answer 1

No. Take $$X=Y \sim \mathcal{N}(0,1), \quad f(x)=-x.$$ It has the requested properties (since also $-X \sim \mathcal{N}(0,1)$), but $$ \text{Cov}(X,Y)=\text{Var}(X)=E[X^2] - E[X]^2 \neq -E[X^2] + E[X]^2 = \text{Cov}(-X,Y).$$