Solve for $\varphi_0$ $$\varphi_{n+1}(a)=\int_{-\infty}^{\infty}\varphi_{n}(x)\varphi_0(x-a)dx$$
$$\lim_{n\to\infty}\varphi_n(a)=e^{-|a|}$$
Solve for  $\varphi_0:\mathbb{R}\to\mathbb{R}$
This is not a homework. What I have tried might be terribly wrong and misleading.

Thank you Yves, following your Fourier transformation I could immediately get:
$$\varphi_0=\delta(x)$$
Well, then further problems arise as $\lim_{n\to\infty}\varphi_n(x)=e^{-|x|}$ is only one of the possibilities.
For example, take $\varphi_0=\delta(x)=\lim_{b\to\infty}e^{-x^2/2^b}$, then $\varphi_\infty(x)$ become the form of  $e^{-x^2}$

Strictly in Math, Dirac delta is an operator rather than a function.
Not every $\delta[f(x)]$ works here.
Could you please guide me through the complexity?
 A: Take the anzatz $\varphi_0(x)=ae^{\alpha |x|}$:
Then:
$$
\varphi_{1}(x)=\int_{-\infty}^{\infty} \varphi_{0}(y) \varphi_{0}(y-x)dy\\
=\int_{-\infty}^{\infty} ae^{\alpha |y|} ae^{\alpha |y-x|}dy\\
$$
If $x\ge 0$:
$$
\varphi_{1}(x)=\int_{0}^{\infty} a^2e^{\alpha (2y+x)} dy+\int_0^x a^2e^{\alpha (y-y+x)} dy+\int_{x}^{\infty} a^2e^{\alpha (2y-x)} dy\\
= \left. \frac {a^2}{2\alpha}e^{\alpha (2y+x)} \right|_{0}^{\infty}
+ \left. a^2 e^{\alpha x}y    \right|_0^x
+ \left. \frac {a^2}{2\alpha}e^{\alpha (2y-x)} \right|_{x}^{\infty}
$$
All limits exist and vanish only for $\alpha=-|\alpha|<0$, then:
$$
= \frac {a^2}{2|\alpha|}e^{-|\alpha| x} 
+  {a^2}x e^{-|\alpha| x} 
+ \frac {a^2}{2|\alpha|}e^{-|\alpha| x} \\
\varphi_{1}(x)= {a^2}x e^{-|\alpha| x}  +\frac {a^2}{|\alpha|}e^{-|\alpha| x} \\
$$
If $x\le 0$, the integrals are the same than for $|x|$, so:
$$
\varphi_{1}(x)= {a^2}|x| e^{-|\alpha| |x|}  +\frac {a^2}{|\alpha|}e^{-|\alpha| |x|} \\
$$
For the anzatz be able to converge:
$$
\frac{a^2}{|\alpha|}=a \to a=|\alpha|>0
$$
and:
$$
\varphi_{1}(x)= {a^2}|x| e^{-a |x|}  +ae^{-a |x|} \\
$$
The first term vanishes in the next iterations, as you can realize (under work...!).
