Effect of test charge on electric field intensity Obviously on the macroscopic scale, having a test charge being a proton produces no noticeable change in the electric field produced by a charged body.
However, on the microscopic level this would definitely make a difference. So, does a positive test charge (+e) cause the apparent electric field intensity (produced by a change say +10e) to be larger or smaller than the actual value?
What would happen if the source charge was negative?
Does it have to do with the attraction or repulsion between unlike and like charges?
 A: "Does it have to do with the attraction or repulsion between unlike and like charges?" Yes.
The electric field always goes out from positive charges and in towards negative charges. The magnitude of the electric field of the +10e charge is independent of the test charge.
\begin{equation}
E=\frac{Q}{4\pi\epsilon r^2}
\end{equation}
However, the direction of the electric field changes near the vicinity of the test charges. If the test charge is positive then the electric field lines would not meet since the field lines of both the particles are coming out of them and are thus in opposite directions. But if the test charge was negative then the field lines for one particle would be going inside while for the other particle it would be going outside. Since the field lines are going in the same direction (from positive charge towards negative charge) they would meet.
This will happen even if charge placed is $+10e$ while the test charge is only $1e$. Most of the region would be dominated by the field lines of $10e$ however, at some point, close to the $1e$ charge, they will cancel each other out and will not meet- causing the charges to repel.
