Step 1: Recall the potential energy.
For the $n^{th}$ orbit, $U=\frac{-27.2}{n^2}$ eV, always negative.
Step 2: Compare the two orbits.
At $n=1$, $U=-27.2$ eV. At $n=5$, $U=\frac{-27.2}{25}\approx-1.09$ eV.
Step 3: Decide the trend.
$-1.09$ is greater (less negative) than $-27.2$, so the potential energy goes up as the electron moves outward. \[ \boxed{\text{is increased}} \]