What if you define something like
$$T(x[n]) = \begin{cases}x[n] & \text{ if }x[0] \neq x[1] \\
0 & \text{ if }x[0] = x[1]\end{cases}$$
In other words, ##T## maps the sequence ##\{x[n]\}## to itself if ##x[0] \neq x[1]##, and it maps ##\{x[n]\}## to the zero sequence if ##x[0] = x[1]##.
Verification that ##T## is homogeneous is straightforward.
Case 1: ##c = 0##. Then clearly ##cx[0] = cx[1]##, so ##T(cx[n]) = 0 = cT(x[n])##.
Case 2: ##c \neq 0## and ##x[0] \neq x[1]##. Then ##cx[0] \neq cx[1]##, so ##T(cx[n]) = cx[n] = cT(x[n])##.
Case 3: ##c \neq 0## and ##x[0] = x[1]##. Then ##cx[0] = cx[1]##, so ##T(cx[n]) = 0 = cT(x[n])##.
On the other hand, ##T## is not additive. For example, let
$$x_1[n] = \begin{cases}1 & \text{ if }n = 0 \\
0 & \text{ if }n \neq 0\end{cases}$$
$$x_2[n] = \begin{cases}1 & \text{ if }n = 1 \\
0 & \text{ if }n \neq 1\end{cases}$$
Then ##T## maps ##\{x_1[n]\}## to itself, and ##\{x_2[n]\}## to itself, but ##T## maps ##\{x_1[n] + x_2[n]\}## to the zero sequence.