Let $X_1,X_2...$ be uniformly bounded random variables and $C_1,C_2>0$ such that $$\left\lvert \operatorname{Cov}(X_i,X_j)\right\rvert < C_1\cdot e^{-C_2|i-j|}$$
Does this imply exponential decay of $\operatorname{Cov}(X_i^2,X_j^2) $ as well? (i.e., are there $C_3,C_4$ such that $$\left\lvert \operatorname{Cov}(X_i^2,X_j^2)\right\rvert < C_3\cdot e^{-C_4|i-j|})$$
In general this doesn't have to be true for $$\operatorname{Cov}(f(X_i),f(X_j)) $$ for any $f$, but intuitively I would expect it for a polynomial (and specifically $f(x)=x^2$).
In general no: let $\left(\varepsilon_i\right)_{i\geqslant 1}$ be an i.i.d. sequence of random variables such that $\Pr(\varepsilon_i=1)=\Pr(\varepsilon_i=-1)=1/2$ and let $X$ be a random variable independent of the sequence $\left(\varepsilon_i\right)_{i\geqslant 1}$ and such that $X^2$ is not constant. Finally, let $X_i=\varepsilon_i X$. Then for each $i\geqslant 1$, by independence of $X$ and $\varepsilon_i$, $\mathbb E\left[X_i\right]=\mathbb E\left[\varepsilon_i\right]\mathbb E\left[X\right]=0$. Therefore, for $i\neq j$, $$ \operatorname{Cov}\left(X_i,X_j\right)=\mathbb E\left[X_iX_j\right]= \mathbb E\left[ \varepsilon_i\varepsilon_j X^2\right]=\mathbb E\left[ \varepsilon_i\varepsilon_j \right]\mathbb E\left[ X^2\right]=0. $$ Moreover, $$ \operatorname{Cov}\left(X_i^2,X_j^2\right)=\mathbb E\left[X_i^2X_j^2\right] - \mathbb E\left[X_i^2\right]\mathbb E\left[X_j^2\right] =\mathbb E\left[X^4\right]-\left(\mathbb E\left[X^2\right]\right)^2\neq 0. $$