I don't think this is true in general. The point is that that there are non-isomorphic groups with equivalent categories of representations; since the category of representations together with the fiber functor determines the group, this gives a counterexample.

This happens under the following circumstances; the following construction was first given, I believe, in Giraud's book on non-abelian cohomology.

Suppose that $G$ is an affine algebraic group over $\mathop{\rm Spec} K$ and $P \to \mathop{\rm Spec}K$ is a $G$-torsor. Call $H$ the group scheme of automorphisms of $P$ as a torsor; then $P$ becomes an $(H, G)$-bitorsor, that is, admits commuting actions of $G$ on the right and of $H$ on the left, and is a torsor for both. Conversely, if $P \to \mathop{\rm Spec}K$ is an $(H, G)$-bitorsor, then $H$ is the automorphism group scheme of $P$ as a $G$-torsor.

Then the categories of representations of $G$ and $H$ are isomorphic. This follows, essentially, from descent theory; if $V$ is a representation of $G$, then the quotient $(P \times_{\mathop{\rm Spec}K} V)/G$ is a vector space on $K$ with an action of $H$; the inverse functor is obtained by exchanging $G$ and $H$ (and right and left actions).

My favorite example of this is the following: if $q$ and $q'$ are non-degenerate quadratic forms in $n$ variable, the orthogonal groups $\mathrm O(q)$ and $\mathrm O(q')$ have equivalent categories of representations (although they are not isomorphic, in general). The bitorsor is the functor of isometries of $q$ and $q'$.

On the other hand, if $K$ is algebraically closed the fiber functors are indeed isomorphic; if memory serves me well, this is in Deligne's paper in Grothendieck's Festschrift, but I don't have it here and can't check right now.