I have a particular function that for even numbers $m$ obeys the following equation:
$$f_{m,n}\left(\frac{2}{m}-x\right)=(-1)^nf_{m,n}(x)$$
Now when I put in odd values for $m$ and plot the function, they are not equal, but there appears to be a reflectional symmetry along the vertical line $x=\frac{1}{m}$. I am having trouble with the appropriate transformation of reflectional symmetry along an axis different than the $y$-axis. I'm thinking it would be $x\mapsto{-(x-1/m)}$. Is this correct?
To reflect across $x = k$, just translate this line to go through the origin, reflect across the $y$-axis, and then translate back:
$$x \longmapsto x - k \longmapsto -(x - k) \longmapsto -(x - k) + k = 2k - x.$$
So you're right about an axis of symmetry $x = \frac{1}{m}$, but your functional equation actually spells out the transformation $x \mapsto \frac{2}{m} - x$ when $n$ is even.
You'll have rotational symmetry about $\left(\frac{1}{m}, 0\right)$ when $n$ is odd.