You can think of the even and odd parts of a function as decomposing the function into its +/-1 eigencomponents with respect to the operator f(x) -> f(-x).
You can think of the exponential Fourier series of a function as a way to decompose the function into its {..., -2, -1, 0, 1, 2, ...} eigencomponents with respect to the operator f(x) -> f'(x).
I like this from a representation theory perspective. The cyclic group of order two, C_2, is the set {1,x} with x^2=1 (it's reasonable to just use {1,-1} with the group operation being real-number multiplication). Let V be the set of continuous function R->R, and then let's define a linear representation where phi_1 is the identity operator on V and phi_x is the operator you describe, phi_x(f)(c)=f(-c). (The only two symmetries of a line which fix the origin are the identity and flipping, which is what this representation is representing.)
The group C_2 is known to have exactly two irreducible representations (the positive and the negative representations), so V decomposes into (at most) two subrepresentations (that is, there are two subspaces of V which are closed under the group action). Using the characters of C_2, we get two projection operators: (\phi_1+\phi_x)/2 and (\phi_1-\phi_x)/2. Examining what these do, they decompose a function into the even and odd parts, respectively!
This idea can be extended to the circle group for the Fourier transform.
Representations are able to capture a bit more than eigencomponent. Where an eigencomponent requires that the action be strictly scaling, components from representation can have more complicated actions. For instance, if you have a representaiton of the dihedral group of symmetries of a triangle, then there will be projections which will give you 1) the +1 eigencomponent, 2) the -1 eigencomponent associated with flipping the triangle over, and 3) the 2-dimensional component which faithfully represents the symmetries of the triangle (i.e., the one already mentioned when describing the dihedral group).
It's funny that you mention this; I was about to type a similar comment! Thinking of the Fourier transform in terms of group theory seems like it would make it more complicated, but it actually makes the fundamental underlying concept simpler to understand.
One can perform the Fourier transform over an arbitrary, compact, non-commutative group G via f(g) = ∑ dᵏ⋅tr(f̂ᵏ⋅ρᵏ(g)), where the sum is from k = 0 to k = ∞, and k indexes the unitary, irreducible representations ρᵏ of G. dᵏ is the dimension of the kth representation. f̂ᵏ is the (matrix) Fourier coefficient of the kth irreducible representation and is computed as f̂ᵏ = ∫ f(g)⋅ρᵏ(g⁻¹) dμ(g), where μ is a Haar measure on G such that ∫ dμ(g) = 1. Note that since the group representations are unitary, ρᵏ(g⁻¹) = ρᵏ(g)ᴴ.
For commutative groups, all of the ρᵏ are one-dimensional, and so the sums and integrals are over scalar values. As you mention, for the circle group the above expression reduces to the "conventional" equation for the Fourier transform.
One can think of the group Fourier transform as decomposing a nonlinear function over a group into a linear combination of orthonormal functions such that cutting off the sum at the kth term provides the best MSE approximation to the function, i.e., f̂ᵏ = argmin ∫ [tr(ĉᵏ⋅ρᵏ(g)) - f(g)]² dμ(g), where the minimization is over ĉᵏ.
You can think of the exponential Fourier series of a function as a way to decompose the function into its {..., -2, -1, 0, 1, 2, ...} eigencomponents with respect to the operator f(x) -> f'(x).