Can you determine if a correlation exists independent from outliers without conducting an experiment? Let's say A correlates with B. But, A is correlated with C, D, E, and F which also correlate with B.
Could you determine if A's correlation with B is solely due to the fact that C, D, E, and F correlate with B, or is that logically impossible without running an experiment?
If adding A to a multiple regression with the other variables greatly increases Adjusted R-squared would that be enough evidence to suggest that A has an effect on B independent from the other variables? If so, is there a test you could perform?
 A: Let's simplify the question slightly, and consider only one correlate. So $A$ correlates with $B$, as well as $C$ which also correlates with $B.$ There are a number of different scenarios, corresponding to whether $C$ is a confounder or not. Here is a causal graph for the true confounder:

$A$ causally influences $B$, but $C$ causally influences both $A$ and $B.$ In the new causal revolution lingo, $C$ sets up a back-door path from $A$ to $B.$ In this case, if you were to do a multiple regression
$$B=aA+cC,$$
you would condition on $C,$ and thus correctly identify the causal effect of $A$ on $B.$ In this case, the correlation between $A$ and $B$ might go down, but it shouldn't disappear.
Now let's take a completely different scenario, called the mediator:

Here there is no back-door path from $A$ to $B,$ and hence $C$ is not a confounder. It is called a mediator, instead. If you were to do the same regression as before, you would actually NOT get the true causal effect of $A$ on $B,$ because $C$ should not be conditioned on in this case. Now it is a theorem (Theorem 1.2.8 on page 19 in Pearl's Causality: Models, Reasoning, and Inference, 2nd. Ed. (2009)) that two directed acyclic graphs (such as we have here) "are observationally equivalent if and only if they have the same skeletons and the same sets of $v$-structures, that is, two converging arrows whose tails are not connected by an arrow." These two graphs clearly have the same skeleton (remove the directions on the arrows to get an undirected graph - that is your skeleton), and there are no $v$-structures. So these two graphs are observationally equivalent. That means you must resort to experimental manipulation to distinguish which of these two graphs model your scenario better.
So the final answer is that you must experiment.
I would add that these two graphs by no means exhaust the possibilities. You might have a simple $A\to C\to B,$ for example. Or you could have $A\leftarrow C\to B,$ or $A\to C\leftarrow B.$ That last does have a $v$-structure, so you could distinguish it from the other models I've mentioned in this paragraph.
I do not see that the presence or absence of outliers changes any of this analysis.
