Find closed form of $f(x)= \frac{1}{x} \sum_{i=1}^{k} f(x+i)$

Under some regularity assumptions, let $\hat{f}$ be the Fourier transform of $f$, $$ \hat{f}(\xi) = \int_{-\infty}^{\infty} f(x)\ e^{-2\pi j x \xi}\,dx.$$ Here I note $j^2=-1$. You can re-write your equation as $xf(x) = \sum_{i=1}^kf(x+i).$ Take Fourier transform of both sides, $$\frac{1}{2\pi j}\hat{f}'(\xi) = \sum_{i=1}^k e^{2\pi j i \xi} \hat{f}(\xi).$$ Now we have an equation of the form $\hat{f}'(\xi) = h(\xi) \hat{f}(\xi)$ with $ h(\xi) = 2\pi j\sum_{i=1}^k e^{2\pi j i \xi}.$ Any primitive of $h$ is of the form $H(\xi) = C + \sum_{i=1}^k \frac{e^{2\pi j i \xi}}{i}$ for any constant $C\in\mathbb{C}$. Thus $ \hat{f}(\xi) = e^{H(\xi)}$ and $f$ is its reverse Fourier transform: $$ f(x) = \int_{-\infty}^{\infty} \hat{f}(\xi)\ e^{2\pi j x \xi}\,d\xi.$$ In other words, here is a closed form solution to your original equation: $$ f(x) = c\int_{-\infty}^{\infty} \exp{}\left(\sum_{i=1}^k \frac{e^{2\pi j i \xi}}{i}+ 2\pi j x \xi\right)\,d\xi,$$ with $c$ an arbitrary constant.

edit: my mistake, of course it does not satisfy the necessary regularity assumptions... So my solution is not valid.

Rewriting:

$$f(x)= \frac{1}{x} \sum_{n=1}^{k} f(x+n)$$

under the form:

$$xf(x)= \sum_{n=1}^{k} f(x+n)$$

we get, using Laplace Transform:

$$-F'(s)= \sum_{n=1}^{k} e^{ns}F(s)$$

giving a differential equation:

$$\dfrac{F'(s)}{F(s)}+ \sum_{n=1}^{k} e^{ns}=0$$

that can be integrated without difficulty:

$$\ln(F(s))+\sum_{n=1}^{k} \frac{1}{n}e^{ns}+C=0$$

$$F(s)=\exp(-\sum_{n=1}^{k} \frac{1}{n}e^{ns}+C)$$

Having $F(s)$, it remains to come back into the original space with variable $x$. This is possible (but not very simple) using the Mellin inversion formula.

But there is a big problem (I fact, we meet cousin problems with Marin). I realize here that in the simple case $k=1$ we cannot retrieve $\Gamma$ function because... $\Gamma$ function has no Laplace Transform (due to its too rapid increase).

Conclusion: It is necessary to use functional transforms that are neither Fourier nor Laplace...

Without additional hypotheses, there is an infinite set of solutions to this problem. To see this, consider $k=1$. We can build a solution $f$ to the problem $$f(x) = \frac 1 x f(x+1)\tag{1}$$

Let $h$ be an arbitrary function defined on $(0, 1]$.

Start by defining $f(x)=h(x)$ for $x\in(0,1]$.

Then construct $f$ on $(1, 2]$ by using $(1)$: For $x\in (1, 2]$ define $$f(x) = (x-1)f(x-1)=(x-1)h(x-1)$$ You can repeat the process to construct $f$ on $(2, 3]$. For $x\in (2,3]$, define $$f(x) = (x-1)f(x-1)=(x-1)(x-2)h(x-2)$$ and so on... to construct $f$ on each interval $(n, n+1]$ with $n\in\mathbb N^*$.

For negative numbers, the same applies: You can construct $f$ on $(-1, 0)$ by using $(1)$: For with $x\in (-1, 0)$, $$f(x) = \frac 1 x f(x+1) = \frac 1 x h(x+1)$$

Then you can repeat the process for $x\in(-2, -1)$: $$f(x)=\frac 1 x f(x+1) = \frac 1 x \frac 1 {x+1} h(x+2)$$ and so on...

This way you can construct $f$ on the entire real line (except for non-positive integers) from an arbitrary $h$.

To obtain that the $\Gamma$ function and its multiples are the unique solution, you need additional hypotheses, such as $\ln f$ being convex. See the Bohr-Mollerup theorem.