Because Fourier transform is a linear transformation in the Hilbert space of square-normalizable curves $L^2$, this is essentially an eigenvalue problem. You are looking for a function that doesn't change under Fourier transform. Beside the Gaussian function, which is very famously an eigenfunction of the Fourier transform, there is a whole class of them that also satisfy this eigenvalue condition of $Ff=cf$. They are essentially a Gaussian times a Hermite polynomial:
http://en.wikipedia.org/wiki/Hermite_functions#Hermite_functions_as_eigenfunctions_of_the_Fourier_transform
On the other hand, $F^4=I$ is universal. If you observe the Fourier transform integral, you can see that the Fourier transform and inverse Fourier transform have an opposite sign in the imaginary exponent. This is why $F$ is not exactly $F^{-1}$ but it's close. $F^2$ flips the function (reverses time) because of this mismatch in sign, but reversing time twice brings you back again, which is why $F^4=I$ works for any function, not just the eigenfunctions. This is related to $i^4=1$: $F^4=I$ tells you that the only eigenvalues of $F$ are $c\in\{\pm 1, \pm i\}$ (try $F^4 f=F^3 (cf)=c^4 f=f$ which gives you $c^4=1$). Of course, this means, that the Fourier space is highly degenerate. A linear combination of any number of eigenfunctions with the same eigenvalue is also an eigenfunction. You can use the Hermite functions (they sequentially cycle through eigenvalues $i^n$, so take every fourth) as a basis spanning each of the four subspaces: you get an infinite family of eigenfunctions for each eigenvalue.
Be warned that this depends on the normalization of the Fourier transform. You need the unitary version. If you take the regular definition (the one used in physics and engineering), you get additional factors of $\sqrt{2\pi}^n$, which the transform blows up or extinguishes the amplitude of the function, and is not cyclic anymore.
You will notice that you can split any function into 4 components with eigenvalues $\{1,i,-1,-i\}$ by doing this:
$$\frac{1}{4}(1+F+F^2+F^3)f=f_1$$
$$\frac{1}{4}(1-iF-F^2+iF^3)f=f_i$$
$$\frac{1}{4}(1-F+F^2-F^3)f=f_{-1}$$
$$\frac{1}{4}(1+iF-F^2-iF^3)f=f_{-i}$$
where
$$Ff_1=f_1\quad Ff_i=if_i\quad Ff_{-1}=-f_{-1}\quad Ff_{-i}=-if_{-i}$$
Each of these components has a very specific symmetry. In particular, the $\pm i$ are odd and $\pm 1$ are even.
You can take any arbitrary function and use it to generate a function that is preserved under Fourier transform. This I think is the most general answer I can give you.
Your question means nothing. Work on
$$\text{I}\Pi(x) = \sum_{k=-\infty}^\infty \delta(x-k) = 1+2 \sum_{n=1}^\infty \cos(2\pi n x)$$
where the Fourier series on the right converges only is in the sense of distributions, that is, for every $\varphi \in C_c^\infty$ with (*) compact support $[a,b]$ :
$$\langle \text{I}\Pi, \varphi \rangle = \int_{-\infty}^\infty \text{I}\Pi(x) \varphi(x) \, dx = \sum_{k \in \mathbb{Z} \cap [a,b]} \varphi(k)$$
$$=\lim_{N \to \infty} \langle 1+2 \sum_{n=1}^N \cos(2\pi n x), \varphi \rangle = \lim_{N \to \infty} \int_{-\infty}^\infty (1+2 \sum_{n=1}^N \cos(2\pi n x)) \varphi(x) \, dx$$
So what I mean is : there is no Fourier series for the Dirac delta, there is only a Fourier series for the Dirac comb $\text{I}\Pi(x)$.
And read a course on : the Fourier series and the Fourier transform, on the distributions, on some complex analysis and the Laplace/Mellin transform.
(*) Since $\text{I}\Pi(x)$ is tempered distribution of order $1$, you can extend $\langle \text{I}\Pi,\varphi \rangle$ to any $\varphi(x)$ continuous (at $x \in\mathbb{Z}$) and with compact support, or decreasing fast enough at $x \to \infty$.
For example, it is perfectly true that $\langle \text{I}\Pi(x) ,x^{-s}\Lambda(\lfloor x+1/2 \rfloor) \rangle = \sum_{n=1}^\infty n^{-s}\Lambda(n) = \frac{-\zeta'(s)}{\zeta(s)}$ for $Re(s) > 1$, but it doesn't mean that
$$\frac{-\zeta'(s)}{\zeta(s)} = \lim_{N\to \infty} \int_{1/2}^\infty x^{-s} \Lambda(\lfloor x+1/2 \rfloor)(1+2\sum_{n=1}^N \cos(2\pi n x)) \, dx$$
and this is all the point of the theory of distributions (and the theory of the Riemann zeta function) to study those kind of things.
Best Answer
In short a tempered distribution $T$ is uniquely defined by its pairing with gaussians : $$T_{a} = e^{-\pi a^2 x^2} (T \ast \frac1a e^{-\pi x^2/a^2})$$ Note $T_{a}$ is $C^\infty \cap L^1$
(when $a \to 0$, $\frac1a e^{-\pi x^2/a^2} \to \delta$ and $T \ast \frac1a e^{-\pi x^2/a^2} \to T$ in the sense of distributions).
Given a tempered distribution, compute the Fourier transform $\hat{T}_{a} = FT[T_a] \in C^\infty \cap L^1$. and let $a \to 0$, you obtain $\hat{T} = \lim_{a \to 0} \hat{T}_a$ (again limit in the sense of distributions) and $\hat{T}_a = \frac1a e^{-\pi x^2/a^2} \ast (e^{-\pi \xi^2 a^2} \hat{T})$, which, if true for one $a$, is true for every $a$.