Is it possible to solve an equation look like this? $$
\int_{-\infty}^{\infty} f(x+t) \cdot g(t)dt=f(x), \forall x \in \mathbb{R}
$$
Any necessary regularity conditions can be assumed to hold.
I was wondering if there exists a generic method to solve an equation like this.
 A: Despite the ambiguities of the question, there are some possibly-interesting things that can be said.
First, with or without the change of variables (as @ThomasAndrews points out) to rewrite the equation as a convolution equation $h*g=g$, we can certainly view the left-hand side as an integral operator $T=T_g$ (using auxiliary function $g$) applied to $f$, and the equation requires that $f$ be an eigenvector for $T_g$, with eigenvalue $\lambda=1$.
If $T_gf=\lambda_g\cdot f$ is required for every $g$ in a sufficiently large class of functions (for example, compactly-supported and smooth, that is, test functions), the condition is (provably) equivalent to requiring that $f$ be an eigenfunction for translations, and the associated differential equation is $f'=c\cdot f$. So $f(x)=e^{cx}$ for some constant $c$.
None of these functions is in $L^2$, etc. But the eigenvalues are easy to determine:
$$\int e^{c(x+t)}\,g(t)\;dt \;=\; \Big(\int e^{ct}\,g(t)\;dt\Big)\cdot e^{cx}
$$
Second, with the rewrite to have a convolution equation $f*g=\lambda\cdot f$, if we assume that $f,g$ have Fourier transforms (in various possible senses), then $\widehat{f}\cdot \widehat{g}=\lambda\cdot \widehat{f}$. That is, (assuming both Fourier transforms have pointwise senses), $\widehat{g}(x)=\lambda$ on the support of $\widehat{f}$, is a necessary and sufficient condition.
For this to be possible for a wide range of $g$, it is necessary that the support of $\widehat{f}$ not be the whole line. If $f$ itself was compactly supported (and continuous, for example), then $\widehat{f}$ extends to an entire function, so cannot vanish on any set of positive measure. In that case, $\widehat{g}(x)=\lambda$, and $g$ was $\lambda\cdot \delta$ with Dirac $\delta$.
... and the equation holds for all continuous $f$, for $g=\lambda\cdot\delta$.
EDIT: as in @user3716267's comment/question, it may not be sooo common to understand precisely how being an eigenfunction under translation leads to the differential equation. Yes, we know (by the Mean Value Theorem) that translation invariance of $f$ gives $f$ constant, so $f'=0 = 0\cdot f$.
In fact, $f(x+y)=\lambda_y\cdot f(x)$ for all $x,y$ implies that
$$
{\partial\over\partial y}\Big|_{y=0} f(x+y) \;=\; {\partial\over \partial y}\Big|_{y=0}(\lambda_y)\cdot f(x)
$$
from which $f'(x)={\partial\over \partial y}\Big|_{y=0}(\lambda_y)\cdot f(x)$. Yes, some (of various possible) hypotheses are necessary to make this argument rigorous. A variant of it does just reduce to the Mean Value Theorem version for $f$ translation-invariant, by considering $f(x)e^{-\lambda x}$... But this heuristic is perhaps clearer as a suggestion of what is true.
