Let $(X_1, ..., X_n), (Y_1, ... , Y_n)$ be random variables.
$X_i$ has the same distribution as $Y_i$ for all $i$.
$\forall i, j: Cov(X_i, X_j) = Cov(Y_i, Y_j)$
Do $(X_1, ..., X_n)$ and $(Y_1, .., Y_n)$ have the same distribution? Are there other things you can say about the distribution of the two?
Edit: Does anything chance if we assume $Cov(X_i, X_j) > 0$
No, as far as I know this is not the case in general. There is a counterexample for $n=2$, that is related to this example in Wikipedia:
$(X_1,X_2)$ are jointly Gaussian with variance $\sigma$ and vanishing covariance (i.e., $X_1$ and $X_2$ are uncorrelated and independent).
$Y_1$ is Gaussian with variance $\sigma^2$.
$Y_2=WY_1$, where $W=1$ w.p. 0.5 and $W=-1$ w.p. 0.5. Hence, $Y_2$ is also Gaussian with variance $\sigma^2$, and uncorrelated with $Y_1$ (but not independent).
It follows that $X_i$ and $Y_i$ have the same distribution and that the covariance $Cov(X_1,X_2)=Cov(Y_1,Y_2)=0$. Still, $Y_1$ and $Y_2$ are not independent, hence the joint distributions of the two vectors differ.