Ireland

To say that the graph of $f$ is symmetric about the line $x=a$ is to mean that $f(x_1)=f(x_2)$ whenever $x_1,x_2$ are real numbers whose sum is $2a$. Hence, by hypothesis, $$f\left(\frac{1}{4}-x\right)=f\left(\frac{1}{4}+x\right)\text{and}f\left(\frac{3}{4}-x\right)=f\left(\frac{3}{4}+x\right)\text{for all }x\in \mathbb{R}.$$ Letting $y=\frac{1}{4}-x$, we deduce from the first condition that $f(y)=f(\frac{1}{2}-y)$ for all $y\in \mathbb{R}$. Letting $y=\frac{3}{4}-x$ in the second relation, we see that $f(y)=f(\frac{3}{2}-y)$ for all $y\in \mathbb{R}$. Therefore, $$f\left(\frac{1}{2}-y\right)=f\left(\frac{3}{2}-y\right)\text{for all }y\in \mathbb{R}.$$ Hence, letting $x=\frac{1}{2}-y$ we see that $f(x)=f(1+x)$ for all $x\in \mathbb{R}$, as desired.

Consider the function $$f(x)={\sin }^2(2\pi x).$$ Then, for all $x\in \mathbb{R}$, $$\begin{array}{rl}f\left(\frac{1}{4}-x\right)& ={\sin }^2\left(2\pi \left(\frac{1}{4}-x\right)\right)\\ & ={\left(\sin \left(\frac{\pi }{2}\right)\cos (2\pi x)-\cos \left(\frac{\pi }{2}\right)\sin (2\pi x)\right)}^2\\ & ={\cos }^2(2\pi x)=f\left(\frac{1}{4}+x\right).\end{array}$$

$$\begin{array}{rl}f\left(\frac{3}{4}-x\right)& ={\sin }^2\left(2\pi \left(\frac{3}{4}-x\right)\right)\\ & ={\left(\sin \left(\frac{3\pi }{2}\right)\cos (2\pi x)-\cos \left(\frac{3\pi }{2}\right)\sin (2\pi x)\right)}^2\\ & ={\cos }^2(2\pi x)=f\left(\frac{3}{4}+x\right).\end{array}$$ Since it's nonnegative and not a constant function, this $f$ does the job.