What is the dipole moment direction in the nitrosonium ion?
In non-neutral species the dipole moment in calculations is dependent on the choice of the co-ordinate system. For this reason the origin is chosen to be the centre of mass. With $r=107.8~\pu{pm}$ on RMP2/def2-QZVPP previously determined by a geometry optimisation. The program used for the computations is Gaussian 09 Rev. D.01 and used input files are attached at the bottom of the post.

I am loosely following the same procedure/ conclusions as in Gernot Frenking, Christoph Loschen, Andreas Krapp, Stefan Fau,  and Steven H. Strauss, J. Comp. Chem., 2007, 28 (1), 117-126. (I believe it is available free of charge.) since $\ce{NO+}$ is isoelectronic with $\ce{CO}$.

According to the definition of the dipole moment
$$\huge\ominus \overset{\mathbf{p}}{\longrightarrow}\oplus$$
positive values in the following table indicate that the negative end is at the nitrogen atom.

\begin{array}{lr}\hline
\text{Orbital} &amp; \text{Bond Moment}           \\\hline
\text{MO 1 } (\sigma)            &amp;   -4.8336  \\
\text{MO 2 } (\sigma)            &amp;    5.5232  \\
\text{MO 3 } (\sigma)            &amp;   -0.7891  \\
\text{MO 4 } (\sigma)            &amp;   -3.0061  \\
\text{MO 5 } (\pi   )            &amp;   -1.1642  \\
\text{MO 6 } (\pi   )            &amp;   -1.1642  \\
\text{MO 7 } (\sigma)            &amp;    4.7383  \\\hline
\text{MO 8 } (\pi, \text{virt.}) &amp;   (2.3831) \\
\text{MO 9 } (\pi, \text{virt.}) &amp;   (2.3831) \\\hline
\text{Total } \sum_i^7 \text{MO }i               &amp;   -0.6956  \\\hline
\end{array}

While we see the same polarisation of the molecular orbitals as in carbon monoxide, the magnitude differs and the HOMO of $\ce{NO+}$ cannot compensate for the polarisation of the electron density towards the oxygen.





Appendix

Base calculation for $\ce{NO+}$ after previous optimisation, E(RHF)=-128.972090453 input file:


%chk=mp2qzvpp.chk
#p MP2/def2QZVPP
scf(verytight)
symmetry(loose)
gfinput gfoldprint iop(6/7=3)

base

+1 1
N   0.0 0.0 -0.575312
O   0.0 0.0  0.503398



Analysis calculation input file for MO 1


%oldchk=mp2qzvpp.chk
#p MP2/def2QZVPP
geom=check
guess=(read,only,alter)
symmetry(loose)
gfinput gfoldprint iop(6/7=3)

analysis

+13 1




(Both empty lines are necessary.)

Analysis calculation input file for MO 2, etc


%oldchk=mp2qzvpp.chk
#p MP2/def2QZVPP
geom=check
guess=(read,only,alter)
symmetry(loose)

analysis

+13 1

2,1



The last line needs to contain the complete reordering procedure. The MO to be analysed needs to be in first place. Therefore for MO 3 it needs to read 3,1,2, for 4 4,1,2,3, etc.