If "self opposite" means "isomorphic to its opposite ring," then no.
If you take an infinite dimensional $V$, then it is known that the ring of linear transformations written on the left is right self-injective but not left self-injective. So it cannot be isomorphic to its opposite ring.
In the finite dimensional case, the answer is yes, of course, by using the transposition map.
If "self opposite" means "isomorphic to its opposite ring," then no.
If you take an infinite dimensional $V$, then it is known that the ring of linear transformations written on the left is right self-injective but not left self-injective. So it cannot be isomorphic to its opposite ring.
In the finite dimensional case, the answer is yes, of course, by using the transposition map.