Denote $f(0) = c$. We show that $f(x) = c$ for every real number $x$. Fix a real number $a$ arbitrarily; by assumption, the function $g(x) = f(x + \frac{a}{2})$ is either an even or odd function. If this $g$ is even then
$$f(a) = f\left(\frac{a}{2} + \frac{a}{2}\right) = g\left(\frac{a}{2}\right) = g\left(-\frac{a}{2}\right) = f\left(-\frac{a}{2} + \frac{a}{2}\right) = f(0) = c.$$
Analogously in the case of odd $g$ we obtain
$$f(a) = f\left(\frac{a}{2} + \frac{a}{2}\right) = g\left(\frac{a}{2}\right) = -g\left(-\frac{a}{2}\right) = -f\left(-\frac{a}{2} + \frac{a}{2}\right) = -f(0) = -c.$$
Thus if $g$ defined as in the problem is even for all choices of $a$ then $f(x) = c$ for every real number $x$. If $g(x) = f(x + a)$ is an odd function for some $a$ then $f(a) = f(0 + a) = g(0) = 0$. But by the above, $f(a) = c$ or $f(a) = -c$; hence $c = 0$. This implies that, for every real number $x$, $f(x) = 0 = c$.
On the other hand, if $f$ is a constant function then $g(x) = f(x + a)$ is the same constant function for every $a$. Constant functions are even. Hence constant functions satisfy the conditions of the problem.