设等差数列\(\left\{ {{a_n}} \right\}\)的公差为d，当\(n \geqslant 2\)时，\({a_n} - {a_{n - 1}} = d\).
对于A，\({a_{n + 1}} + 3 - \left( {{a_n} + 3} \right) = {a_{n + 1}} - {a_n} = d\)，为常数，
因此\(\left\{ {{a_n} + 3} \right\}\)是等差数列；故A正确
对于B，\(a_{n + 1}^2 - a_n^2 = d\left( {{a_{n + 1}} + {a_n}} \right) = d\left[ {2{a_1} + \left( {2n - 1} \right)d} \right]\)，不为常数，
因此\(\left\{ {a_n^2} \right\}\)不是等差数列；故B错误
对于C，\(\left( {{a_{n + 2}} + {a_{n + 1}}} \right) - \left( {{a_{n + 1}} + {a_n}} \right) = {a_{n + 2}} - {a_n} = 2d\)，为常数，
因此\(\left\{ {{a_{n + 1}} + {a_n}} \right\}\)是等差数列；故C正确
对于D，\(2{a_{n + 1}} + \left( {n + 1} \right) - \left( {2{a_n} + n} \right) = 2\left( {{a_{n + 1}} - {a_n}} \right) + 1 = 2d + 1\)，为常数，
因此\(\left\{ {2{a_n} + n} \right\}\)是等差数列，故D正确
故选：ACD.
