No you can't. But, it is true that for any function (that's a characterization of the conditional expectation).

You can use the above equality with : it gives . Since additionaly , this justifies your second equality.Therefore i conclude that:

Cov(X, E[Y/X])

= E[(X - E[X])(E[Y/X] - E[E[Y/X]])]

= E[(X - E[X])(Y- E[Y])]

= Cov(X, Y)

Thanks in advance!