Rational normal curves and tangent lines Let $C,\Gamma\subset\mathbb{P}^n$ be degree $n$ rational normal curves in $\mathbb{P}^n$, such that for any $p\in C$ the tangent line $T_pC$ of $C$ at $p$ is tangent to $\Gamma$ as well. This means that there exists a point $q\in\Gamma$ such that $T_q\Gamma = T_pC$.
Clearly, if $n = 2$ this means that $C$ and $\Gamma$ are two conics having the same dual conic i.e. $C^{*} = \Gamma^{*}$, and therefore $C = \Gamma$.
Is it true that this property (any tangent line of $C$ is tangent to $\Gamma$ as well) forces $C = \Gamma$ ?
 A: Edit. This answer has been edited to address positive characteristic, as alerted by @FelipeVoloch.  The result holds except in characteristic $2$.  In characteristic $2$, the result fails.
Let $k$ be an algebraically closed field.  Let $C$ be a smooth, connected $k$-curve.  Let $V$ be a $k$-vector space of dimension $n+1$, so that $\mathbb{P}V$ is $k$-isomorphic to $\mathbb{P}^n_k$.  In order to state the result carefully, recall a few definitions related to linear systems on smooth curves.
Definition 1. A $V$-linear system on $C$ is a pair $(\mathcal{L},\phi_0)$ of an invertible $\mathcal{O}_C$-module, $\mathcal{L}$, and a homomorphism of $\mathcal{O}_C$-modules, $$\phi_0:V^\vee\otimes_k \mathcal{O}_C \to \mathcal{L}.$$  
Definition 2. For every integer $r\geq 0$, the functor of principal parts of order $\leq r$ is the exact $k$-linear functor, $$\mathcal{P}^r_{C/k}:\mathcal{O}_C-\underline{\text{mod}} \to \mathcal{O}_C-\underline{\text{mod}},\ \ \mathcal{E} \mapsto \text{pr}_{1,*}(\text{pr}_2^*(\mathcal{E})\otimes \mathcal{O}_{C\times_k C}/\mathcal{I}_{\Delta}^{r+1}),$$ with respect to the projections, $$\text{pr}_i:C\times_{\text{Spec}\ k} C \to C, i=1,2,$$ and the ideal sheaf of the diagonal,  $\mathcal{I}_{\Delta}\subset \mathcal{O}_{C\times_k C}.$ 
There is a filtration of $\mathcal{O}_{C\times_k C}/\mathcal{I}_\Delta^{r+1}$ by the powers of $\mathcal{I}_\Delta$.  The associated graded pieces are isomorphic to $\Delta_*\Omega_{C/k}^{\otimes r}$, and these are flat for the projections.  Thus, the induced filtration on $\mathcal{P}^r_{C/k}(\mathcal{E})$ gives a natural sequence of short exact sequences, $$ 0\to \mathcal{E}\otimes_{\mathcal{O}_C}\Omega_{C/k}^{\otimes r} \xrightarrow{q_r(\mathcal{E})} \mathcal{P}^r_{C/k}(\mathcal{E}) \xrightarrow{p^r_{r-1}(\mathcal{E})} \mathcal{P}^{r-1}_{C/k}(\mathcal{E}) \to 0,$$ for $r=1,2,\dots$  This is discussed in the exercises of ACGH, in the first volume of Illusie's "Complexe Cotangent et Deformations", etc.  In particular, adjointness of pullback and pushforward combined with the natural isomorphisms $$\text{pr}_2^*(V^\vee\otimes_k \mathcal{O}_C) \cong V^\vee\otimes_k \mathcal{O}_{C\times_k C} \cong \text{pr}_1^*(V^\vee\otimes_k \mathcal{O}_C),$$ define a natural splitting $$\sigma_r:V^\vee\otimes_k \mathcal{O}_C \to \mathcal{P}^r_{C/k}(V^\vee\otimes_k \mathcal{O}_C)$$ of the surjective composition $p^r_0(\mathcal{E}) = p^1_0(\mathcal{E})\circ \dots \circ p^r_{r-1}(\mathcal{E})$.  
Definition 3. For every $V$-linear system on $C$, for every integer $r=0,\dots,n$, the associated system of  degree-$r$ principal parts is the composition, $$V^\vee\otimes_k \mathcal{O}_C \xrightarrow{\sigma_r} \mathcal{P}^r_{C/k}(V^\vee\otimes_k \mathcal{O}_C) \xrightarrow{\mathcal{P}^r(\phi_0)} \mathcal{P}^r_{C/k}(\mathcal{L}).$$
For every open subscheme $U$ of $C$, thelinear system $(\mathcal{L},\phi_0)$ is $r$-jet globally generated on $U$ if $\phi_r|_U$ is surjective on $U$.  In this case, the associated morphism for osculating $r$-planes is the unique $k$-morphism, $$f_{U,r}:U\to \text{Grass}(r+1,V)$$  such that the pullback by $f_{U,r}$ of the universal rank $r+1$ quotient sheaf of $V^\vee\otimes_k \mathcal{O}_{\text{Grass}(r,V)}$ equals $\phi_r|_U$.  
If $U$ is dense in $C$, then by the valuative criterion of properness, the morphism $f_{U,r}$ extends uniquely to a $k$-morphism $f_r$ on all of $C$, the pullback by $f_r$ of the universal quotient is a subsheaf of $\mathcal{P}^r_{C/k}(\mathcal{L})$ whose cokernel is a torsion coherent sheaf.  The ramification points of the linear system are precisely the points in the support of these torsion sheaves, and the ramification sequence at each ramification point determines the length of the stalk of this torsion sheaf.  Let $q$ be a $k$-point of $C$.
Hypothesis 4  The stalk of $\phi_0$ at $q$ is injective, $$(\phi_0)_q:V^\vee \hookrightarrow \mathcal{L}_q. $$
There is a filtration of $\mathcal{L}_q$ by powers of the maximal ideal $\mathfrak{m}_{C,q}$, and the inverse image is a filtration of $V^\vee$ by $k$-subspaces.  Under the hypothesis, this is a bounded filtration.  Denoting by $t$ a generator of the principal ideal $\mathfrak{m}_{C,q}$, there exists an ordered $k$-basis of $V^\vee$, $$(x_0,\dots,x_n):k^{\oplus n} \xrightarrow{\cong} V^\vee,$$ there exists a sequence of nonnegative integers, $$0\leq \alpha_0\leq \alpha_1 \leq \dots \leq \alpha_n,$$ and there exists a sequence $(u_0,u_1,\dots,u_n)$ of invertible elements, $$u_i \in 1 + \mathfrak{m}_{C,q},$$ such that for every $i=0,\dots,n$, the germ $$\phi_{q,i}:= (\phi_0^*x_i)_q \in \mathcal{L}_q,$$ equals $t^{i+\alpha_i}u_i$.
Definition 5.  The ramification sequence of $(\mathcal{L},\phi_0)$ at $q$ is the sequence of nonnegative integers $(\alpha_0,\dots,\alpha_n)$. The ramification index at $q$ is the sum $\alpha_0+\dots + \alpha_n$.
Proposition 6. The linear system is globally generated on an open neighborhood of $q$ if and only if it is $0$-jet globally generated on an open neighborhood of $q$ if and only if $\alpha_0$ equals $0$. In this case, the $k$-morphism $f_0$ is unramified on an open neighborhood of $q$ if and only if the linear system is $1$-jet globally generated on an open neighborhood of $q$ if and only if $\alpha_1$ also equals $0$.  For $r\geq 2$, if the characteristic of $k$ does not divide $r!$, then $(\phi_r)_q$ is surjective if and only if the following germ is a generator of the rank $1$ module, $$\text{Wronski}^r(\phi_{q,0},\dots,\phi_{q,r};t):= $$ $$\phi_{q,0}\wedge \frac{d\phi_{q,1}}{dt} \wedge \dots \wedge \frac{d^r\phi_{q,r}}{dt^r} \cdot (dt)^{r(r+1)/2} \in \bigwedge^{r+1}_{\mathcal{O}_C}\mathcal{P}^r_{C/k}(\mathcal{L})_q,$$ and this holds if and only if $\alpha_0=\dots = \alpha_r=0.$  If the characteristic divides $r!$, then the linear system is not $r$-jet globally generated.
Proof.  This is the usual argument with the Wronskian of the germs $\phi_{q,0},\dots,\phi_{q,r}$.  In particular, modulo $\mathfrak{m}_{C,q}\bigwedge^{r+1}\mathcal{P}^r_{C/k}(\mathcal{L})_q$, the Wronskian element is congruent to $r! t^{\alpha_0+\dots+\alpha_r}\phi_{q,0}^{r+1} dt^{r(r+1)/2}$.  This is nonzero if and only if both the characteristic does not divide $r!$, and $\alpha_0+\dots+\alpha_r$ equals $0$. QED
Nota bene.  If the characteristic equals $2$, then always $(\phi_2)_q$ is not surjective.  
Proposition 7. Assume that characteristic of $k$ is not $2$, and assume that $\alpha_2$ equals $0$.  Then for the $1$-dimensional $k$-subspace $L\subset V$ parameterized by $f_0(q)$, resp. for the $2$-dimensional $k$-subspace $S\subset V$ parameterized by $f_1(q)$, the subspace $L$ equals the unique $k$-subspace of $S$ that is annihilated by every element of the image of the derivative $d_q f_1$ considered as a $k$-linear transformation in $T_{[S]}\text{Grass}(2,V)=\text{Hom}_k(S,V/S).$
Proof  This is easiest to check with respect to the basis above.  Up to redefining the uniformizing element, assume that $\phi_{q,1}$ equals $t\phi_{q,0}$.  Let $(e_0,\dots,e_n)$ be the ordered $k$-basis for $V$ that is dual to the ordered $k$-basis $(x_0,\dots,x_n)$ for $V^\vee$.  Then $L$ is $\text{span}(e_0)$ and $S$ is $\text{span}(e_0,e_1)$.  The derivative of $f_1$ at $q$ sends the basis vector $d/dt$ of $T_q C$ to the $k$-linear transformation, $$d_q f_1(d/dt):S\to V/S,$$ $$ e_0 \mapsto -\sum_{r=2}^n \left((r+\alpha_r-1)u_r + t(du_r/dt)\right)t^{r+\alpha_r}e_r,$$ $$e_1 \mapsto \sum_{r=2}^n \left((r+\alpha_r)u_r + t(du_r/dt)\right)t^{r+\alpha_r-1}e_r.$$  Modulo $\mathfrak{m}_{C,q}$, the span of $e_0$ maps to $0$.  Under the hypothesis that $\alpha_2$ equals $0$ and that the characterstic does not equal $2$, the image of $e_1$ is nonzero modulo $\mathfrak{m}_{C,q}$.  Thus, the annihilator is precisely the spane of $e_0$, i.e., it is $L$. QED
Corollary 8.  For germs $(C,q)$ and $(C',q')$ of pointed smooth $k$-curves, for $V$-linear systems $(\mathcal{L},\phi_0)$, resp. $(\mathcal{L}',\phi'_0)$, on these germs, if the characterstic does not equal $2$, if each linear system is $2$-jet globally generated, and if the germ of the image of the unramified morphism $f_1$ at $f_1(q)$ equals the germ of the image of the unramified morphism $f'_1$ at $f'_1(q')$ as closed curves in $\text{Grass}(2,V)$, then the germ of the image of the unramified morphism $f_0$ at $f_0(q)$ equals the germ of the image of the unramified morphism $f'_0$ at $f'_0(q')$ as closed curves in $\mathbb{P}(V)$.
In particular, so long as the characteristic does not equal $2$, then germs of rational normal curves are $2$-jet globally generated.  Thus, if $1$-pointed rational normal curves in $\mathbb{P}^n$ have equal germs in $\text{Grass}(2,V)$, then the pointed curves in $\mathbb{P}^n$ are equal.  
More counterexamples in characteristic $2$.  As Felipe Voloch alerted, if $\text{char}(k)$ equals $2$, then the rational normal curve is not uniquely determined by the set of tangent lines.  Here is a classification of counterexamples in $\mathbb{P}^n$ with $n$ equal to  $2m+1$, an odd integer $\geq 3$.  The classification in even dimension is similar.
Consider the scroll that is the image of the following closed immersion of $k$-schemes, $$\tau:\mathbb{P}^1\times \mathbb{P}^1 \to \mathbb{P}^{2m+1},$$ $$\tau^*[x_0,x_1,\dots,x_{2e},x_{2e+1},\dots,x_{2m},x_{2m+1}] = $$ $$[s^mu,s^mv,\dots,s^{m-e}t^eu,s^{m-e}t^ev,\dots,t^mu,t^mv],$$ where $(s,t)$, resp. $(u,v)$, form an ordered $k$-basis for $H^0(\mathbb{P}^1\times \mathbb{P}^1,\text{pr}_1^*\mathcal{O}(1))$, resp. for $H^0(\mathbb{P}^1\times \mathbb{P}^1,\text{pr}_2^*\mathcal{O}(1)).$  
For every $k$-point $[a,b,c,d]$ of $\textbf{PGL}_2$, i.e., $ad-bc\neq 0$, let $C_{a,b,c,d}$ be the curve in $\mathbb{P}^1\times \mathbb{P}^1$ with defining equation, $$g(s,t,u,v) = (au^2+bv^2)t-(cu^2+dv^2)s = 0.$$  As $[a,b,c,d]$ varies in $\textbf{PGL}_2$, these are all distinct closed curves in $\mathbb{P}^1\times \mathbb{P}^1$. 
For every $k$-point $\lambda \in \mathbb{P}^1$, the fiber $F_{\lambda}:=\text{pr}_1^{-1}(\{\lambda\})$ is an effective Cartier divisor in $\mathbb{P}^1\times \mathbb{P}^1$ that is tangent to $C_{a,b,c,d}$.  Thus, the image $L_{\lambda} = \tau(F_{\lambda})$ is a tangent line to the rational normal curve $\tau(C_{a,b,c,d})$.  Therefore, independent of $(a,b,c,d)\in \textbf{PGL}_2(k)$, the set of tangent lines of the rational normal curve $\tau(C_{a,b,c,d})$ equals the set of all lines $\tau(\text{pr}_1^{-1}(\{\lambda\}))$.  Since the union of these lines equals the image of $\tau$, the only rational normal curves in $\mathbb{P}^{2m+1}$ having this collection of tangent lines is one of the curves $\tau(C_{a,b,c,d})$ as above.  
In summary,the set of tangent lines to the rational normal curve $C\subset\mathbb{P}^n$ of degree $2n+1$ sweeps out the unique rational surface scroll $\Sigma$ of degree $2m$ and Hirzebruch type $\Sigma \cong \mathbb{P}^1\times \mathbb{P}^1$ containing  $C$ as a divisor of bidegree $(1,2)$ and whose non-isomorphic projection $$C\hookrightarrow \Sigma\twoheadrightarrow \mathbb{P}^1,$$ is purely inseparable.  The set of such divisors in $\Sigma$ forms a faithful $\text{Aut}(\mathbb{P}^1)$-orbit.  
A: I want to suggest a simpler argument than Jason's, but I am not sure whether it works in positive characteristic or not.
Assume $n > 2$.
Consider the surface $S(C)$ swept by tangent lines to $C$. It is easy to see that the surface is not normal, its normalization is isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$, its map to $\mathbb{P}^n$ is given by (an incomplete) linear system $\mathcal{O}(1,n-1)$, such that $C$ is the image of the diagonal in $\mathbb{P}^1 \times \mathbb{P}^1$ and is equal to the singular locus of the surface.
Now, assume that every tangent to $C$ is also tangent to $\Gamma$. Then $S(C) \subset S(\Gamma)$, and since both surfaces are irreducible, we have $S(C) = S(\Gamma)$. But then $C = Sing(S(C)) = Sing(S(\Gamma)) = \Gamma$.
