Center of the group of invertible bounded linear operators with bounded inverse on Normed linear spaces Let $X$ be an Normed linear space. Let $Iso(X)$ be the set of all invertible bounded linear operators on $X$ with bounded inverse. If $T \in Iso(X)$ is such that $T \circ S=S \circ T, \forall S \in Iso (X)$, then is it true that $T=kI$ for some scalar $k$ ?
If $X$ is Banach space, then the claim is true, which I can show as follows: Let $0\ne v \in X$. Pick $0\ne L_v \in X^*=\mathcal L(X,F)$ ($F$ is the underlying field) such that $||L_v||<\dfrac {1}{2||v||}$. Fix $a\in X$ with $L_v(a)\ne 0$. Define $S:X \to X$ as $S_v(x)=L_v(x)v,\forall x \in X$. Then $||S_v||\le ||L_v||||v||<1/2$. Then $I-S_v \in Iso(X)$ (this is where I need $X$ to be Banach). Let $T \in Iso(X)$ which commutes with every member of $Iso(X)$. Then we have in particular $T((I-S_v)(a))=(I-S_v)(T(a))$ i.e. $T(a-S_v(a))=T(a)-S_v(T(a))$ i.e. $T(a)-T(S_v(a))=T(a)-S_v(T(a))$ i.e. $T(L_v(a)v)=L_v(T(a))v$ i.e. $L_v(a)T(v)=L_v(T(a))v$ i.e. $T(v)=k_v v$ for some scalar $k_v \in F$. And now it can be easily shown that $k_v=k_w,\forall 0\ne v,w \in X$, hence $T$ is a scalar multiple of identity. 
I don't know what happens if $X$ is not Banach. 
Please help. 
 A: EDIT: First attempt at a proof was wrong, I hope this works now. Apparently this needs the Hahn-Banach theorem in order to show that in any normed vector space there exist continuous projections onto one-dimensional subspaces.

This is true in general. First show that $T$ preserves every one-dimensional subspace. Assume not. Then there exist linearly independent $v$ and $w$ with $T(v) = w$. By the Hahn-Banach theorem there exists a bounded linear functional $\phi:X \to F$ with $\phi(v) = 0$, $\phi(w) = 1$ and $\|\phi(u)\| \le \| u \|$ for all $u \in X$. Now $P:X \to X$ defined by $P(u) = \phi(u) w$ is a bounded projection operator onto the subspace generated by $w$, with $v$ in the kernel of this projection $P$.
Define operators $S_a = I + aP$ for $a \in F$. Then each $S_a$ is a bounded operator, and $S_a S_b = (I+aP)(I+bP) = I + (a+b+ab)P = S_{a+b+ab} = S_{(a+1)(b+1)-1}$, since $P^2 = P$. In particular, if $(a+1)(b+1) = 1$, e.g., if $a = 1$ and $b=-1/2$, then $S_a S_b = S_b S_a = S_0 = I$, so every $S_a$ for $a \ne -1$ is an invertible bounded linear operator with bounded inverse, satisfying $S_a(v) = v$ and $S_a(w) = (1+a)w$. Picking $a=1$, we get that $w = T(v) = T(S_1(v)) = S_1(T(v)) = S_1(w) = 2w$, which is a contradiction because $w \ne 0$.
Now if $v$ and $w$ are two linearly independent vectors, we know that $T(v) = \lambda v$ and $T(w) = \mu w$. If $\lambda \ne \mu$, then $T(v+w) = \lambda v + \mu w = \lambda (v+w) + (\mu - \lambda)w$ is linearly independent of $v+w$, contradicting the fact that $T$ preserves the subspace generated by $v+w$. This shows that there exists a constant $\lambda$ with $T(v) = \lambda v$ for all $v$.
