SAT formula and satisfying assignment yield another different satisfying assignment
Your problem is NP-complete. Let's call it NextSAT$(F,\tau)$: accept iff $\tau$ is a satisfying assignment of $F$ and there exists a satisfying assignment of $F$ different from $\tau$.

NextSAT is clearly in NP. We now show that SAT $\leq$ NextSAT. 

Let $F$ be a CNF on variables $X = \{x_1,\ldots,x_n\}$. If $F$ is the empty CNF, it is obviously satisfiable and we can thus pick any satisfiable instance of NextSAT for the reduction. 

Assume now that $F$ has at least one clause. Choose $\tau : X \rightarrow \{0,1\}$ that does not satisfy $F$. It is easy to find by simply choosing a clause of $F$ and negating all its literals. Let $$G = F \lor (\bigwedge_{i \leq n} \ell_i) = \bigwedge_{C \in F} \bigwedge_{i \leq n} C \lor \ell_i$$ where $\ell_i = x_i$ if $\tau(x_i) = 1$ and $\ell_i = \neg x_i$ otherwise. We see that $G$ can be written as a CNF of size polynomial in the size of $F$. 

Now, clearly, the satisfying assignments of $G$ is the disjoint union of the satisfying assignments of $F$ and $\tau$ since $\tau$ is the only satisfying assignment of $\bigwedge_{i\leq n} \ell_i$ and does not satisfy $F$. Thus, $F$ is satisfiable if and only if there exists a satisfying assignment of $G$ different from $\tau$. That is, if and only if NextSAT$(G,\tau)$ returns $1$.