As Thierry and Qiaochu have pointed out, it's not clear what exactly the question means. Even so, maybe it's possible to say something helpful.

The question asks

Can one always have that $N(t) = M^t$ holds for some non-singular $M$?

This doesn't quite make sense, so I'll interpret it as

Must there exist a non-singular matrix $M$ such that $N(t) = M^t$ for all $t$?

I'm guessing this was the intention.

The answer is probably no. (I say "probably" because the hypotheses on $N$ weren't stated precisely.) It's even false for $n = 1$. That is, there exist functions $N: \mathbb{R} \to (0, \infty)$ satisfying
$$
N(s + t) = N(s) N(t)
$$
for all $s, t \in \mathbb{R}$, but not of the form $N(t) = M^t$ for any $M > 0$. The existence of such functions requires the axiom of choice.

This is a problem going back to Cauchy. It's equivalent to consider the equation $f(s + t) = f(s) + f(t)$ (by putting $f(s) = \log N(s)$), and this is what's usually called "Cauchy's functional equation". You can find information about it here and here, for example.

Maybe you're happy to assume that $N(t)$ is continuous in $t$, in which case there's a genuine question about matrices to be answered. But in the absence of precise hypotheses, I don't know what you're happy to assume.