Functional Square Root

[Leo.W]

On the topic of function composition. Is the successor function the only function f(x) such that f(f(x)) = f(x)+1?

The equation shows f is one Abel function of itself, and thus also the superfunction of itself.
...
this shows f(x)=x+1 is the only solution.

F(x) = ∞ is also a solution.

What about Σ n = 0 to ∞ f^n(x)/n! = exp(f(x)). What is f?
What about Σ n = 0 to ∞ f^n(x)/n! = f(exp(x))?

Well if you insist, then there will be also a solution to the equation x=x+1, which is x=∞, that'll make sense to some extent, but only considering finite solutions as we usually do, then f(x)=x+1 is the only. 

For these 2 equations, if you're asking finite analytic solution, I guess you can solve it by the fixed point method. For example:

Assume L is one solution of f(L)=L, then f^n(L)=L, gives L*e=exp(L), thus we know all fixed points of f(x) has to solve L*e=exp(L), for example L=1, f(1)=1.
Then we take derivatives of both sides at x=1, gives e^(f'(1))=e^(f(1))f'(1), so f'(1) satisfies ex=e^x, let's again assume f'(1)=1, then we can solve by induction for f''(1), f'''(1) and on.
But if you really take attempts, you'll find that f'(1) cannot be 1, which would make the sum diverge (seen easily from the second derivative), to make it converge, you have to find a solution to ex=e^x where the absolute value of x mustnot exceed 1, which is impossible.
And this shows the fixed point of f cannot be 1.

If the fixed point of f is any L not 1, we'll have L*e=exp(L) and exp(f'(L))=exp(L)f'(L)=e*L*f'(L), again we have to find a solution of exp(x)=e*L*x whose absolute value remains smaller than 1. As I computed,
\[L=-W_1(-\tfrac{1}{e})\approx{3.08884-7.46149i}, s=f'(L)=-W_0\bigg(\frac{1}{eW_1(-\tfrac{1}{e})}\bigg)\approx{0.0158263+0.0434912i}\]
Then we can work out the series:
\[\begin{matrix}L\approx{3.088843015613044-7.461489285654255i},\\f(z+L)-L\approx(0.01582633039817652 +
    0.04349124710235049 i) z\\+ (0.0008417750386726426-
    0.0006753346652964609 i) z^2\\- (0.00002833079697273823 +
    0.00001790952001363610 i) z^3\\- (2.664193985539843*10^{-7} -
    1.1390607179533432*10^{-6} i) z^4\\+ (4.329246420079456*10^{-8} -
    4.56979799629351*10^{-9} i) z^5\\- (7.19187320031943*10^{-10} +
    1.526422873318864*10^{-9} i) z^6\\- (4.803646824452359*10^{-11} -
    4.709364757070538*10^{-11} i) z^7\\+ (2.453010990405934*10^{-12} +
    1.215725690054748*10^{-12} i) z^8\\+ (1.410524507975909*10^{-14} -
    1.1230584812952190*10^{-13} i) z^9+O(z^{10})\end{matrix}\]
for 16 digits precision in decimal

You can solve the other in the similar way, but I don't see f(x) has a closed form expression.