Why Lagrange multipliers? Your maximization problem can be solved in a rather straightforward manner using some standard tricks in matrix theory. Let $e=\frac1{\sqrt{n}}(1,\ldots,1)^T\in\mathbb{R}^n$ and $X=(x_1,\ldots,x_n)\in M_{m,n}(\mathbb{R})$ (hence $X$ is "tall" and $X^T$ is "wide"). The problem can be formulated as maximizing
$$
f(X)=\textrm{tr}(X^TAX)+ne^TX^TAXe=\textrm{tr}\left((I+nee^T)X^TAX\right)
$$
subject to the constraint $X^TX=\frac{n+1}{n}I-ee^T$.
The eigenvalues of $X^TX$ are $\frac{n+1}{n}$ (with multiplicities $n-1$) and $\frac{1}{n}$ (with $e$ being an eigenvector). Pick any orthogonal matrix $V$ whose last column is $e$. Then every $X$ that satisfies $X^TX=\frac{n+1}{n}I-ee^T$ can be written as $X=U\Sigma V^T$, where $U$ is some $m\times m$ orthogonal matrix, $\Sigma$ is the $m\times n$ (tall) diagonal matrix with diagonal $\left(\sqrt{\frac{n+1}{n}},\ldots,\sqrt{\frac{n+1}{n}},\sqrt{\frac{1}{n}}\right)$, and $V$ is an $n\times n$ orthogonal matrix. Now let $e_n=(0,\ldots,0,1)^T\in\mathbb{R}^n$. Then
$$
\begin{align}
\Sigma V^T(I+nee^T)V\Sigma^T &= \Sigma\left[I+n(V^Te)(e^TV)\right]\Sigma^T \\
&= \Sigma(I+ne_ne_n^T)\Sigma^T \\
&=\textrm{diag}\left(\underbrace{\frac{n+1}{n},\ldots,\frac{n+1}{n}}_{n \textrm{ entries}},\ \underbrace{0,\ldots,0}_{m-n \textrm{ entries}}\right)
= D\quad\textrm{(say)}.
\end{align}
$$
Hence
$$
\begin{align}
f(X)&=\textrm{tr}\left((I+nee^T)X^TAX\right)\\
&=\textrm{tr}\left((I+nee^T)V\Sigma^T U^TAU\Sigma V^T\right)\\
&=\textrm{tr}\left(\Sigma V^T(I+nee^T)V\Sigma^T U^TAU\right)\\
&=\textrm{tr}\left(DU^TAU\right)
\end{align}
$$
and the maximum of $f$ occurs when $U^TAU$ is a diagonal matrix whose diagonal entries are in descending order. Thus the answer follows.