$H=U+pV$, so a change in enthalpy will amount to: $dH=dU+VdP+P dV$. 

If you want $dU=0$ you can impose $dT=0$. Using the equation of state for an ideal gas we get: $V dP+P dV=nRdT$, and because you want $dT=0$ in your process you have $VdP=-PdV$.

Replacing that in $dH$ you get:

$$dH=-2PdV$$ Thus an isothermal expansion results in a change in enthalpy but keeps $U$ unchanged.