Maximize $L^p$ norm over sphere

Proof for the case $p$ odd:

The Lagrange-multiplier equation yields $$ p\lambda_i^{p-1}+a\lambda_i+b=0 $$ for some $a,b \in \mathbb{R}$. If $p>1$ is odd the LHS is a strictly convex function in $\lambda_i$ and can be zero at at most two different real values.

Assume now we have $n_1, n_2 \in \mathbb{N}$ times $\lambda_1, \lambda_2$ then we want to maximize $$ \sum_{i=1}^2 n_i\lambda^p_i $$ with the constraints $$ \sum_{i=1}^2 n_i\lambda^2_i=1, \sum_{i=1}^2 n_i\lambda_i=0, \sum_{i=1}^2 n_i=n $$ or equivalently that $$ n_1=\frac{n\lambda_ 2}{ \lambda_2-\lambda_1},\\ n_2=\frac{n\lambda_ 1}{ \lambda_1-\lambda_2},\\ n\lambda_1 \lambda_2=-1. $$ WLOG $\lambda_1>\lambda_2$. Since $\lambda_1 \lambda_2<0$ we have $\lambda_1>0>\lambda_2$. Because $n_1 \geq1$ we have $(n-1)\lambda_2+\lambda_1\leq 0$ and hence $$ \lambda_1^2\leq -(n-1)\lambda_1\lambda_2 =\frac{n-1}{n}.$$ Hence $$ \lambda_1\leq (n-1)^{1/2}n^{-1/2}, \lambda_2=-1/(n\lambda_1) \leq -(n-1)^{-1/2}n^{-1/2}. $$ We then have $$ \sum_{i=1}^2 n_i\lambda^p_i= \frac{\lambda_1^{p-1}-\lambda_2^{p-1}}{\lambda_1-\lambda_2}\leq \frac{(n-1)^{(p-1)/2}n^{-(p-1)/2} -(n-1)^{-(p-1)/2}n^{-(p-1)/2} }{ (n-1)^{-1/2}n^{1/2}}=((n-1)^{p/2}-(n-1)^{-(p-2)/2})n^{-p/2} $$ Equality holds as conjectured by the OP.

Proof for $p$ even:

If $p$ is even the LHS of the Lagrange multiplier equation is strictly convex on the positive halfline and strictly concave on the negative half line. On both half lines we can have at most two solutions. Furthermore we have in total an odd number of solutions. Hence for $p$ even we will have at most three different valus of $\lambda$. Now if we do something similar as in the case $p$ odd we end up with $$ n_2=\frac{1-\lambda_1\lambda_3 n}{(\lambda_2-\lambda_1)(\lambda_2-\lambda_3)}, ... $$ where WLOG $\lambda_1>\lambda_2>\lambda_3$ . Since $n_2>0$ we have $\lambda_1\lambda_3>0$ Hence all three $\lambda 's$ have the same sign and thus they can not all satisfy the Lagrange multiplier equation.