Aut(Q) = {f(x) = q x | q 是非0的有理数}.
证：不难看出,若f是Q的同态,则
f(0) = f(0) + f(0),从而f(0) = 0.
记f(1) = q,则由数学归纳法易见对自然数f(n) = n q.
f(-n) + f(n) = f(0) = 0,从而
f(-n) = - f(n) = - nq.
又归纳知 n f(x) = f(n x),从而
f(x) = f(n x) / n.（x是任意有理数）
即对有理数m / n,有
f(m / n) = f(m) / n.
于是
f((m/n) * y) = (m/n) * f(y),
对上式记x = m / n,并取定y = 1,则
f(x) = x f(1) = x q.
由f是单同态,则Ker f = {0},从而q不为0.
容易验证当q为有理数时,f 还是满同态,从而是同构.
综上,Q的自同构就只有f(x) = q x（q不等于0）.