What is the electron configuration of $\ce{Fe+}$ cation?
\begin{align} \ce{Fe+} &\!:~ [\ce{Ar}]\mathrm{(3d)^6(4s)^1}\label{chm:1}\tag{1}\\ \ce{Fe+} &\!:~ [\ce{Ar}]\mathrm{(3d)^7(4s)^0}\label{chm:2}\tag{2} \end{align}
What I've been told is that \eqref{chm:2} is correct since the $\mathrm{4s}$ electron jumps to $\mathrm{3d}.$
Which representation is correct?
Isolated in a vacuum, in the absence of external fields, the first configuration is correct - $[\ce{Ar}]\mathrm{(3d)^6(4s)^1}$ is the ground electronic state of the iron(I) cation $\ce{Fe^{+}}$. More specifically, the ground state also has the term symbol $\mathrm{^6D_{9/2}}$.
NIST's Atomic Spectra Database has compiled the ground state electronic configurations of almost all possible ions of almost every known element (including such exotic things as $\ce{U^{58+}}$), obtained from a combination of experimental data and theoretical calculations. For example, search for iron, and you will be presented a table with all ground state electronic configurations, from the neutral atom $\ce{Fe^{0}}$ to the hydrogen-like $\ce{Fe^{25+}}$. You can find a description of the output data here.
I'm sure you'd like a reason for why $\ce{Fe^{+}}$ has the first configuration you suggest and not the second, but there is almost certainly no explanation which is both simple and correct. Unfortunately the predictive power of the aufbau principle is somewhat overstated, even when taking into consideration qualitative corrections. Depending on your notions, there are arguably several dozen "unexpected" electronic configurations, across all possible ions of all elements. The worst offenders are the heavier elements (say, the sixth period and onwards), which are rarely (if ever) covered in early chemistry classes.
