Hello Tom, Perhaps I am still not getting it. I am guessing that you are using parenthesis above for two different things (grouping and functions; functions use a square bracket notation) But, doesn't this little experiment suggest that you will not get a unique solution?
Combining both of your equations:
eq[g_, t_, q_] := (1 + t) g[t, q] == (1 + 1/q) g[1/q, 1/t]
Example arbitrary function of two variables:
ftest1[x_, y_] := (y^2/x)
ftest2[x_, y_] := (x^2/y)
eq[ftest1, t, q]
eq[ftest2, t, q]
eq[ArcTan, t , q]
With[
{equation1 = eq[ftest1, t, q], equation2 = eq[ArcTan, t , q],
equation3 = eq[ftest2, t, q]},
ContourPlot[{equation1, equation2, equation3}, {t, -10, 10}, {q, -10,
10}, ContourStyle -> {{Red, Thickness[0.02], Dashed}, {Blue,
Thickness[0.01]}, {Green}}]
]
Seems to show that there common values of t and q are satisfied simultaneously by three different functions.