If we take the definition of a variety as a reduced integral scheme of finite type over an algebraically closed field $k$, then a variety is in particular a scheme over $k$, so is a scheme $X$ with a morphism $X \to Spec(k)$. As I'm understanding it, more emphasis should be put on the morphism instead of the underlying scheme since we can have non-isomorphic schemes over a base $S$ say $\pi,\eta : X \to S$. Then even if we have a finitely generated reduced $k$-algebra $A$, can we still have non-isomorphic $k$-schemes $\eta, \pi : Spec (A) \to Spec (k)$? It seems to me like if this were the case it would be unexpected, since there is just one $k$ variety called Spec$(A)$, but I don't really have anything to base this on other than the fact that the relative point of view is really never applied to varieties.
This is equivalent to putting two non-isomorphic $k$-algebra structures on $A$, where $k$ is algebraically closed, so I the question I'm really interested in is can we have a ring $A$ and two monomorphisms $i_1,i_2 :k \to A$ so that there is no automorphism $\varphi$ of $A$ with $\varphi \circ i_2 = i_1$?