smc

Let $a$ be the length of the smaller base of isosceles trapezoid $ABCD$, and $1$ be the length of the larger base of the trapezoid. The ratio of the smaller base to the larger base is $\frac{a}{1}=a$. Let point $E$ be the foot of the altitude from $C$ to $\overline{AD}$. Since the larger base of the isosceles trapezoid equals a diagonal, $AC=AD=1$. Since the smaller base equals the altitude, $BC=CE=a$. Since the trapezoid is isosceles, $DE=\frac{1-a}{2}$, so $AE=1-\frac{1-a}{2}=\frac{a+1}{2}$. Using the Pythagorean Theorem on right triangle $ACE$, we have $$a^2+{\left(\frac{a+1}{2}\right)}^2=1$$ Multiplying both sides by $4$ gives $$4a^2+(a+1{)}^2=4$$ Expanding the squared binomial and rearranging gives $$5a^2+2a-3=0$$ This can be factored into $(5a-3)(a+1)=0$. Since a must be positive, $5a+3=0$, so $a=\frac{3}{5}$. The ratio of the smaller base to the larger base is $\boxed{\frac{3}{5}}$.