Are matrix exponentials the only continuous homomorphisms $\mathbb R_+\to\mathcal M_n(\mathbb R)$? Let $A:\mathbb R_+\to M_n(\mathbb R)$ be a homomorphism of the additive semi-group of positive reals into the multiplicative semi-group of real matrices. Matrix exponentials $t\to e^{tB}$ are one class of such homomorphisms. If $A$ is differentiable at $0$, then it must be the matrix exponential $e^{tA'(0)}$, since:
$$\frac{A(t+h)-A(t)}h=A(t)\frac{A(h)-I}h\xrightarrow[h\to0]{} A(t)A'(0)$$
Can this sufficient condition be relaxed further to just continuity, maybe even just continuity at zero? This answer from 2010 seems to imply that that's the case, and the theorem is certainly true when $n=1$ because of the uniqueness of positive real roots.
Note: I personally cannot understand any of the algebra in the question linked to in the last paragraph. More elementary answers would be appreciated.

Since $A$ is continuous and $A(0) = I$, $A(t)$ remains in a small disk centered
at $I$ while $t$ is small.  Let $L$ be a branch of log on this disk.  Then
$LA$ is a continuous (germ of an) additive homomorphism.  It is easier to
see (as in the $1$-dimensional case) that continuity applies differentiability (in fact linearity) for additive homomorphisms.  $LA(t)$ must be $tB$ for some matrix $B$, and exponentiating this gives $A(t)$, both only for small $t$, but $A$ is determined by its restriction to any interval.
