Suppose we have $\{Y_{i}\}_{i=1}^{n}$ that is exchangeable so the joint distribution of this sequence is the same as $\{Y_{\sigma(i)}\}_{i=1}^{n}$, where $\sigma:\{1,...,n\}\rightarrow\{1,...,n\}.$
This implies the following statements (I think):
$P(Y_{i}>y_{1},Y_{j}>y_{2})=P(Y_{i}>y_{2},Y_{j}>y_{1})$ $\forall i,j\in\{1,...,n\},i\neq j$, $y_{1},y_{2}\in\mathcal{Y}_{1}\times\mathcal{Y}_{2}$, where $\mathcal{Y}_{i}$ is the support of $Y_{i}$ for $i\in\{1,...,n\}$. This is because if the whole $\{Y_{i}\}_{i=1}^{n}$ joint distribution is exchangeable, it certainly holds for the distribution of any two variables within the sequence should also be exchangeable.
$\text{Cov}(Y_{i},Y_{j})=\text{Cov}(Y_{\sigma(i)},Y_{\sigma(j)})$ for all $i\neq j$. Because, say, $\text{Cov}(.)$ is just a function of two random variables from the exchangeable joint distribution.
Now here is the question. Would it be wrong to say that statements (2) or (1) imply exchangeability of the whole sequence? My possibly quite crude reasoning for thinking (1) does not imply exchangeability is that 'pairwise' independence does not imply joint independence. Exchangeability seems to have a similar flavor to this. Also if (1) does not imply exchangeability then (2) should not since (1) is stronger than (2).