Here's how to finish without assuming the Sylow 5-subgroup is normal.
(Assume by way of contradiction:) If $n=2$, so $g^{-1} f g = ff$, then: $$g^{-2} f g^2 = g^{-1}( g^{-1} f g ) g = g^{-1}( ff )g = g^{-1}( fg g^{-1}f)g = (g^{-1}fg) (g^{-1}fg) = (ff) (ff) = f^4.$$ Every time you apply $g^{-1}( * )g$, you double the number of $f$, so $g^{-i} f g^i = f^{(2^i)}$.
Now we use $g^5 = 1$. If you apply $g^{-1}( * )g$ five times, you get $g^{-5} f g^5 = 1 f 1 = f$ on the one hand, but $f^{(2^5)} = f^{32} = f^2$ on the other, using the fact that 2 ≡ 32 mod 3.
How can $f = f^2$? It does not, since $f$ has order 3. Assuming $n=2$ gives a contradiction, so $n=1$, and $g^{-1} f g = f$. Multiply on the left by $g$ to get $f g = g f$. The group must be abelian, since all of its generators commute.
Now try this with "5" replaced by "2" and you will still get the Sylow 3-subgroup is normal, but the Sylow 2-subgroup need not be normal. The same argument will get you that both n=1 (cyclic group of order six), and n=2 (dihedral group of order six) are possible.
More precisely, you'll check that $g^{-2} f g^2 = f^{(2^2)} = f^4 = f$ just as it should be.
Congratulations, you have just analyzed some semi-direct products using automorphisms! With cyclic Sylow subgroups it makes a lot of sense. We didn't even need to call "n" an automorphism, nor did we need to call M3∪M5 a semi-direct product.
If the subgroups are not Sylow subgroups, then it is harder to check (1) that they are normal or (2) that M∩N = 1, but there are more general techniques for this.
If the subgroups are not cyclic, then you cannot just use one "n" for the Sylow 3-subgroup. You get a matrix of "m"s and "n"s. Not just any matrix will do, but the ones that do work are called automorphisms.
In other words, the general case of semi-direct products is similar to what you are now doing. Don't give up.