Same Activation Energy for Forward and Backward Reactions
The van't Hoff equation describes how the equilibrium constant changes with temperature: 

$$\ln \frac{K_2}{K_1} = −\Delta H_R^\circ (\frac{1}{T_2} − \frac{1}{T_1})$$

So if the reaction enthalpy is zero for a given temperature interval, the equilibrium constant will not change. This also means that the activation energy in either direction is the same, at least if entropy is not considered (the typical diagram explaining activation energy is for a single molecule with potential energy as the y-axis, enthalpy as the difference in energy between reactant and product, and activation energies as the difference between reactant or product and the activated complex).


  If so, please provide an example.


Take the transfer of protons across a membrane. Equilibrium is reached when the concentrations on either side are equal. A change in temperature will not change the equilibrium constant (which is 1, with a reaction enthalpy of zero).

Another example would be a reaction that turns one enantiomer into the other, e.g. D-alanine to L-alanine. In the absence of chiral co-solvents or solvents or other unusual circumstances, again K = 1 and enthalpy = zero over a range of temperatures.