Is SAT with two "opposite" solutions NP-hard? Here is a variant of the SAT problem in which a satisfying assignment must have additional properties.
Input: A 3-CNF formula $f$ with variables $x_{1\dots k}$.
Output: 
For an assignment $S$ of $x_{1\dots k}$, let $\overline S$ be defined such that $x_i=true$ in $\overline S$ if and only if $x_i=false$ in $S$.
Is there an assignment $S$ such that both $f(S)$ and $f(\overline S)$ hold?

Is this problem still NP-hard?

Examples:


$f=(x_1\lor x_2\lor x_3)\land(x_1\lor x_2\lor \neg x_3)\land(x_1\lor \neg x_2\lor x_3)\land(x_1\lor \neg x_2\lor \neg x_3)$
This requires $x_1=true$ in $S$, but then $x_1=false$ in $\overline S$, so $f(S)$ and $f(\overline S)$ cannot simultaneously hold.
$f=(x_1\lor x_2\lor x_3)\land(x_1\lor x_2\lor \neg x_3)\land(x_1\lor \neg x_2\lor x_3)$
$S=\{x_1:true,~x_2:false,~x_3:false\}$
$\overline S=\{x_1:false,~x_2:true,~x_3:true\}$
Then both $f(S)$ and $f(\overline S)$ hold.
Turning my comment to an answer:
The problem you describe is known as Not All Equal SAT (NAE-SAT), but is phrased differently. A NAE assignment for a CNF-formula $\phi$ over variables is one where in each clause there is at least one false variable and one true variable.

It is easy to see that an assignment is NAE iff its inverse is also NAE.

Showing that NAE-SAT is NP-complete is a well-known exercise, and it can be easily solved by splitting it to two parts.

First, given a 3-CNF formula, we can convert it to a 4-CNF formula by adding a variable $w$ and converting every clause of the form $(x\vee y\vee z)$ to $(x\vee y\vee z\vee w)$.

It is easy to see that the original formula is satisfiable iff the resulting formula is in NAE-SAT.

Then, standard techniques can be used to convert this formula back to 3-CNF, while maintaining the NAE property.