CHARGE BINDING AT THE SECOND COORDINATION SPHERE
In #["Fe"("H"_2"O")_5"NO"]"SO"_4#, recall that the charge on sulfate is #2-#. That is a good starting point. It implies the charge on #["Fe"("H"_2"O")_5"NO"]# is #2+# to balance it out.
LIGAND CHARGES IN THE FIRST COORDINATION SPHERE
We know that water is a neutral, weak-field ligand. Thus it does not contribute to the overall charge.
The tricky part is that #"NO"# could either bind as #"NO"^(+)# (linear, triple-bonded) or #"NO"^(-)# (bent, double-bonded)... and we don't really know which. This is known as a "non-innocent ligand", because it messes with us when we are trying to find the oxidation state of the transition metal.
Suppose that it was the linear #"NO"^(+)#. Then the oxidation state on iron would be #+1#. If it was the bent #"NO"^(-)#, then the oxidation state on iron would be #+3#. Unfortunately, both answers are available.
NARROWING DOWN WHICH OXIDATION STATE IS MORE LIKELY
Consider the electron configuration of iron, then. The atomic number is #26#, so its configuration is #[Ar]3d^6 4s^2#.
Since the sixth #d# electron is paired, it is sensible for iron to have a #color(blue)(+3)# oxidation state by losing two #4s# and one #3d# electron. This gives a #color(blue)([Ar]3d^5)# configuration and a total electron spin of #5*"1/2" = color(blue)("5/2")#.
Although, it could lose one #4s# electron and still be favorable according to Hund's rule for maximizing total electron spin when possible, giving it a #color(blue)([Ar]3d^6 4s^1)# configuration, a #+1# oxidation state, and a total electron spin of #5*"1/2" + ("1/2" + (-"1/2")) = color(blue)("5/2")#.
That is indeed still the same total electron spin, but the #3d# orbitals are higher in energy than the #4s# orbitals for iron, AND there is one set of electrons paired, increasing the "pairing energy" of this configuration relative to the #+3# oxidation state (the energy associated with pairing two electrons is repulsive and thus positive).
If this configuration could be stabilized, then it would be more likely...
STABILIZING THE NORMALLY LESS-LIKELY CONFIGURATION?
If the nitrosyl ligand is #"NO"^(+)#, then it acts as a good #\mathbf(pi)#-acceptor ligand (isoelectronic with #"CO"#), accepting electron density into its antibonding #pi# orbitals from a #d_(xz)# or #d_(yz)# orbital on iron via a phenomenon called "#pi#-backbonding".
That would indeed stabilize the #d# orbitals by decreasing the amount of electron density that needs to get spread around, thus decreasing the "pairing energy" (making pairing more favorable).
On the other hand, #"NO"^(-)#, which is bent, is basically not as suitable a #pi#-acceptor; technically, its antibonding #pi# orbital is not the right "symmetry" to accept electrons from iron's #d_(xz)# or #d_(yz)# orbitals.
So, I would expect the nitrosyl ligand to be #color(blue)("NO"^(+))# and for iron to be the (normally less-likely) #color(blue)(+1)# oxidation state.
