If $F(x) = \exp(x A) = \sum_{i = 0}^\infty \frac{1}{i!} x^i A^i$ where $F(x), A \in \mathbb{R}^{n \times n}$, then \begin{equation} F(x + y) = F(x) F(y) = F(y) F(x) \end{equation} holds.
When is the converse true, i.e. if $F(x + y) = F(x) F(y) = F(y) F(x)$, then there exists a matrix $A$ such that $F(x) = \exp(x A)$?
"When is the converse true?"
One suitable hypothesis on $F(x)$, other than
$F(x + y) = F(x)F(y) = F(y)F(x), \tag{1}$
is that $F(x)$ be differentiable at $x = 0$, that is, that
$F'(0) = \lim_{h \to 0} \dfrac{1}{h} (F(h) - F(0)) \tag{2}$
exist as a well-defined matrix; another is that
$F(0) \;\; \text{is invertible}. \tag{3}$
Indeed, both (2) and (3) are in fact necessary, as we shall see. In the event that (1) and (2) apply, we have for any $x$ and $h \ne 0$
$\dfrac{1}{h}(F(x + h) - F(x)) = \dfrac{1}{h}(F(x)F(h) - F(x)) = \dfrac{1}{h}(F(h)F(x) - F(x)); \tag{4}$
we further note that
$F(x) = F(x + 0) = F(0 + x) = F(0)F(x) \tag{5}$
via (1); thus (4) becomes
$\dfrac{1}{h}(F(x + h) - F(x)) = \dfrac{1}{h}(F(h)F(x) - F(0)F(x)) = \dfrac{1}{h}(F(h) - F(0))F(x); \tag{6}$
now taking the limit as $h \to 0$ yields, using (2),
$F'(x) = \lim_{h \to 0} \dfrac{1}{h}(F(x + h) - F(x)) = \lim_{h \to 0} \dfrac{1}{h} (F(h) - F(0))F(x) = F'(0) F(x); \tag{7}$
we see that $F(x)$ must satisfy the ordinary differential equation
$F'(x) = F'(0) F(x); \tag{8}$
the unique solution to (8) is
$F(x) = F(0) e^{F'(0) x}, \tag{9}$
which may be easily checked by direct differentiation:
$F'(x) = (F(0) e^{F'(0) x})' = F(0)(e^{F'(0) x})' = F(0) F'(0) e^{F'0) x} = F'(0) (F(0)e^{F'(0) x}) = F'(0) F(x); \tag{10}$
in performing the validating computation (10), we have (tacitly) used the fact that
$F'(0) F(0) = F(0) F'(0)\tag{11}$
which follows from (1), (2):
$F(0) F'(0) = F(0) \lim_{h \to 0}\dfrac{1}{h}(F(h) -F(0)) = \lim_{h \to 0} F(0)\dfrac{1}{h}(F(h) - F(0)) = \lim_{h \to 0}\dfrac{1}{h}(F(h) - F(0)) F(0) = F'(0)F(0). \tag{12}$
In (9), we are almost there, needing only to deal with the factor $F(0)$; we have
$(F(0))^2 = F(0) F(0) = F(0 + 0) = F(0); \tag{13}$
here (3) at last comes into play; we further have
$F(0) = F^{-1}(0)(F(0))^2 = F^{-1}(0) F(0) = I, \tag{14}$
whence (9) becomes
$F(x) = e^{F'(0) x}, \tag{15}$
the desired form with $A = F'(0)$.
We have thus seen that conditions (1)-(3) are sufficient for (15) to bind; they are in fact also necessary; (3) follows from (15) by setting $x = 0$:
$F(0) = e^{F'(0) 0} = e^0 = I, \;\; \text{invertible}; \tag{16}$
(2) is such an easy consequence of the (given) series expansion of $F(x) = e^{Ax} = e^{F'(0)x}$ that I leave the details to my readers; finally (1), which given (15) may be written
$e^{A(x + y)} = e^{Ax} e^{Ay} \tag{17}$
is a well-known property of exponential matrices; see my answer to $M,N\in \Bbb R ^{n\times n}$, show that $e^{(M+N)} = e^{M}e^N$ given $MN=NM$.
This question and answer are related to Show solution to character identity.
