Solvable Lie algebra whose nilradical is not characteristic

Question

Say that an ideal in a Lie algebra is characteristic if it is invariant under every derivation of the algebra.

It is well known that the nilradical of a finite-dimensional Lie algebra over a field of characteristic $p > 0$ need not be characteristic, but is there an example of a solvable finite-dimensional Lie algebra with non-characteristic nilradical?

It is clear that the nilradical is invariant under automorphisms (this is also sometimes used a a definition of characteristic ideal). For an ideal in a finite-dimensional Lie algebra in characteristic zero, the latter is equivalent to being characteristic. But this is not the case in positive characteristic.

(Cross-posted from MathSE)

YCor · Answer

The question was somewhat answered by the OP in comments of the MathSE duplicate, so let me copy the relevant parts here for the record:

Did you test all low-dimensional solvable Lie algebras in characteristic ?, classified by Willem de Graaf here, for small ?, e.g., for ?=2 or ?=3 ? – Dietrich Burde Mar 3 '16

I've had a look at these, but there are none amongst the 3-dimensional algebras. I haven't yet examined all of the 4-dimensional ones (...). – David Towers Mar 3 '16

I have found a 4-dimensional example over $mathrm{GF}(2)$. – David Towers Mar 7 '16

Feel free to edit this cw answer if you indeed identify the 4-dimensional solvable Lie algebra over a field of characteristic $2$ that works.

Solvable Lie algebra whose nilradical is not characteristic

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