In M Theory, what does the M5 brane look like? In M theory, the fundamental 1 strings are made out of the compactification of an M2 brane. 
In String Theory: Volume 2, Superstring Theory and Beyond
 By Joseph Polchinski Pg. 204
He states that when there is an open M2 brane (which in this case is a 1 string) it attaches it's ends to an M5 brane.
What does an M5 brane look like?
 A: M2 and M5 branes are fundamental objects of M-theory. You can see this from the superalgebra of $D=11$ SUGRA, which is the low energy limit of M-theory. Moreover M2 and M5 couple respectively electrically and magnetically to the RR 3-form $A_3$ of  $D=11$ SUGRA.
In the compactification limit for the eleventh dimension, from M-theory you get the $IIA$ superstring theory. So there must be some correspondence between the objects of the two theories. 
In particular the $D2$ of $IIA$ correspond to an unwrapped M2:
$$\tau_{D2}=\frac{1}{(2\pi)^2g_s (\alpha ')^\frac{3}{2}}=\frac{M_{11}^3}{(2\pi)^2}\equiv T_{M2}$$
Where $M_{11}=\frac{1}{l_{11}}$ is the eleven dimensional Planck mass. 
As you said, the wrapped M2 is identified with a fundamental string, indeed ($R_{11}=g_s (\alpha ')^\frac{1}{2})$:
$$T_{F1}=\frac{1}{2\pi \alpha '}=2\pi R_{11}  T_{M2}$$
Now the unwrapped M5 correspond to the $NS5$ brane:
$$\tau_{NS5}=\frac{1}{(2\pi)^5g_s^2 (\alpha ')^3}=\frac{M_{11}^6}{(2\pi)^5}\equiv T_{M5}$$
And finally the wrapped M5 gives a $D4$ in $IIA$:
$$\tau_{D4}=\frac{1}{(2\pi)^4g_s (\alpha ')^\frac{5}{2}}=2\pi R_{11}  T_{M5}$$
Reference: Basic Concepts in String Theory, (Blumenhagen, Lust, Theisen), pag. 719. 
