答案：
(1) 解：如答图①，过点$C$作$CD \perp x$轴，交$x$轴于点$D$。
$\because$直线$y = -\frac{4}{3}x + 4$与$x$轴、$y$轴分别交于点$A$，$B$，
$\therefore$当$x = 0$时，$y = 4$，
当$y = 0$时，$-\frac{4}{3}x + 4 = 0$，解得$x = 3$，
$\therefore OA = 3$，$OB = 4$。
$\because \angle BAC = 90^{\circ}$，$CD \perp x$轴，
$\therefore \angle AOB = \angle CDA = 90^{\circ}$，$\angle BAO + \angle CAD = 90^{\circ}$，$\angle ABO + \angle BAO = 90^{\circ}$，
$\therefore \angle CAD = \angle ABO$。
又$\because AC = AB$，$\therefore \triangle AOB \cong \triangle CDA(AAS)$，
$\therefore CD = OA = 3$，$AD = OB = 4$，
$\therefore OD = OA + AD = 7$，$\therefore C$点坐标为$(7,3)$。
(2) 证明：如答图②，在$x$轴上截取$AH = BE$，连接$CH$，$EF$。
由(1)可得$\angle CAH = \angle ABE$，
$\because AC = AB$，$AH = BE$，
$\therefore \triangle CAH \cong \triangle ABE(SAS)$，
$\therefore CH = AE$，$\angle BEA = \angle CHA$。
$\because AE = AF$，$\angle EAF = 90^{\circ}$，
$\therefore \angle EAO + \angle FAO = 90^{\circ}$，
又$\because \angle EAO + \angle AEO = 90^{\circ}$，
$\therefore \angle AEO = \angle FAO$，$\therefore \angle AEB = \angle FAG$，
$\therefore \angle FAG = \angle CHG$。
又$\because CH = AE = AF$，$\angle CGH = \angle FGA$，
$\therefore \triangle FAG \cong \triangle CHG(AAS)$，
$\therefore CG = FG$，即$G$是$FC$的中点。
(3) 解：由平移的性质，设直线$A'B'$的函数表达式为$y = -\frac{4}{3}x + m$，
当$x = 0$时，$y = m$，当$y = 0$时，$x = \frac{3}{4}m$，
$\therefore M$点坐标为$(0,m)$，$A'$点坐标为$(\frac{3}{4}m,0)$。
由(1)知$OA = 3$，$OB = 4$，
$\therefore AB = 5$，$BM = 4 - m$，
当以$A$，$B$，$M$为顶点的三角形是等腰三角形时：
①当$AM = BM$时，$3^{2} + m^{2} = (4 - m)^{2}$，解得$m = \frac{7}{8}$，
此时点$A'$的坐标为$(\frac{21}{32},0)$；
②当$AM = AB$时，$3^{2} + m^{2} = 5^{2}$，解得$m = 4$（与点$B$重合，舍去）或$m = -4$，
此时点$A'$的坐标为$(-3,0)$；
③当$AB = BM$时，$4 - m = 5$，解得$m = -1$，
此时点$A'$的坐标为$(-\frac{3}{4},0)$。
综上，点$A'$的坐标为$(\frac{21}{32},0)$或$(-3,0)$或$(-\frac{3}{4},0)$。