Step 1: Try a small concrete case to build intuition.
Take $n = 2$. Then $\mathbf{1} = (1,1)^T$ and $\mathbf{1}\mathbf{1}^T = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$.
$$M = I_2 - \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 0.5 & 0.5 \\ 0.5 & 0.5 \end{pmatrix} = \begin{pmatrix} 0.5 & -0.5 \\ -0.5 & 0.5 \end{pmatrix}$$
Step 2: Test each property on this concrete M.
$M^T = \begin{pmatrix} 0.5 & -0.5 \\ -0.5 & 0.5 \end{pmatrix} = M$, so $M$ is symmetric, consistent with option (A).
$M^2 = \begin{pmatrix} 0.5 & -0.5 \\ -0.5 & 0.5 \end{pmatrix}\begin{pmatrix} 0.5 & -0.5 \\ -0.5 & 0.5 \end{pmatrix} = \begin{pmatrix} 0.5 & -0.5 \\ -0.5 & 0.5 \end{pmatrix} = M$, not $I_2$, so option (B) fails here.
$\text{Trace}(M) = 0.5 + 0.5 = 1 = n - 1$ (since $n=2$), not $n = 2$, so option (C) fails here too.
Since $M^T = M$ and $M^2 = M$, $M$ satisfies the definition of a projection matrix, supporting option (D).
Step 3: Confirm the pattern holds for general n algebraically.
For any $n$, $\mathbf{1}^T\mathbf{1} = n$ because it sums $n$ ones. Using this:
$$M^2 = I_n - \frac{2}{n}\mathbf{1}\mathbf{1}^T + \frac{1}{n^2}\mathbf{1}(\mathbf{1}^T\mathbf{1})\mathbf{1}^T = I_n - \frac{2}{n}\mathbf{1}\mathbf{1}^T + \frac{1}{n}\mathbf{1}\mathbf{1}^T = I_n - \frac{1}{n}\mathbf{1}\mathbf{1}^T = M$$
so $M$ is always idempotent, matching what the $n=2$ case showed. Also $\text{Trace}(\mathbf{1}\mathbf{1}^T) = \mathbf{1}^T\mathbf{1} = n$, so $\text{Trace}(M) = n - \frac{1}{n}\cdot n = n - 1$ for every $n$, confirming option (C) is always false and option (B) is always false for $n>1$.
Step 4: Conclude.
$M$ is symmetric and idempotent for every $n$, which by definition makes it a projection matrix (specifically, projection onto the hyperplane orthogonal to $\mathbf{1}$, the classic mean-centering operator in statistics). Options (A) and (D) hold in general, while (B) and (C) do not.
$$\boxed{\text{(A) and (D) are correct}}$$