$f:[0,1] \rightarrow [0,1]$ is a strictly decreasing function (i.e. $f(x+\delta)<f(x)\;\forall\;x \in[0,1), \delta>0$), such that $f^{-1}=f$. Further, $f(x)+f(1-x)=1\;\forall\;x\in[0,1]$.
Edit: I earlier said "There is no requirement of continuity". However, while there is no explicit continuity requirement, I think these properties imply continuity as pointed out in the comments.
Does this imply $f$ must be linear?
My approach: I drew the pictures. Due to $f^{-1}=f$, $f$ must look the same when we rotate the page by $90^{o}$ so that the y-axis becomes the x-axis. By strict decreasingness, it must be strictly above $1/2$ for $x<1/2$, equal to $1/2$ for $x=1/2$ (by $f(x)+f(1-x)=1$) and strictly below $1/2$ for $x>1/2$. In addition, due to $f(x)+f(1-x)=1$, when you reflect it around the $y=1/2$ line and then the $x=1/2$ line, you get back the same function. To me it seems this means $f$ in $[1/2,1]$ is a replica of $f$ in $[0,1/2]$ except "dragged down", I think by $1/2$ but I'm not sure. i.e. $f(x)=f(x-1/2)-1/2$ for $x>1/2$, but I'm not sure.
To me it seems the replica property combined with the rotational symmetry property described above, indicates that this can't hold unless $f$ is linear.
Obviously, the above is vague, and I'm not even sure it goes loosely in the right direction. Any help is most appreciated.
So $f$ is strictly decreasing from $[0,1]$ onto $[0,1]$. As it is supposed to be bijective, $f(0)= 1$ and $f(1)=0$. $f$ is also continuous, but we don't need this in what follows.
Let's prove that $f(x) = 1-x$ if we suppose in addition that $f(x) + f(1-x) = 1$ for all $x \in [0,1]$ and $f=f^{-1}$.
Suppose that it exists $a \in (0,1)$ such that $f(a) \lt 1-a$. Then
$$a = f^{-1}(f(a)) \gt f^{-1}(1-a) = f(1-a) = 1 - f(a)$$ and therefore $a + f(a) \gt 1$. A contradiction with the hypothesis $f(a) \lt 1-a$.
We'll get a similar contradiction if we suppose that it exists $a \in (0,1)$ such that $f(a) \gt 1-a$.
Conclusion: the only option is $f(x) = 1-x$, which is indeed linear.